Phy 1 (11)

http://www.rpmauryascienceblog.com/ HEAT AND THERMODYNAMICS PROBLEM Q.1. In a steady state, the temperature at the end A and B of 20 cm long rod AB are 1000C and 00C.

Find the temperature of a point 9 cm from A. Q.2. A thermally insulated vessel containing a gas whose molar mass is M and adiabatic exponent J,

moves with a velocity v. Prove that rise in temperature of the gas resulting from the stoppage of the

vessel is )1(R2

mv2J .

Q.3. In a cyclic process shown in the figure an

ideal gas is adiabatically taken from B to A, the work done on the gas during the process B o A is 30 J,when the gas is taken from A o B the heat absorbed by the gas is 20J. Find out the work done by the gas in the process A o B.

A

B

P

V

Q.4. The initial and final temperature of water as recorded by an observer are (40.6 r0.2)qC and (78.3

r 0.3)qC. Calculate the rise in temperature with proper error limits. Q.5. Figure shows two paths through a gas can be taken from the

state A to the state B. Calculate the work done by the gas in each path.

A

B

C 10 cc

25 cc

10 kPa 30 kPa

Q.6. In steady state, the temperature at the end A and B of 20 cm long rod AB are 1000C and 00C. Find

the temperature of a point 9 cm from A. Q.7. A gas is found to obey P2V = constant. The initial temperature and volume of gas are T0 & V0. If

the gas expands to volume 3V0. Find the final temperature of gas. Q.8. A thermodynamical process is a shown in the

figure with PA = 3 u 105 Pa, vA = 2 u 10

-3 m3, PB = 8 u 10

4 Pa, vC = 5 u 10-3 m3, in the

process AB and BC 600 J and 200 J heat is added to the system, respectively. Find the change in internal energy of the system in the process AC.

P

A

B C

V

PA

PB

VA VC

Q.9. Find the specific heat capacity of mono-atomic ideal gas for thermodynamic process

P = Dv2. Where D is positive constant.
http://www.rpmauryascienceblog.com

http://www.rpmauryascienceblog.com/ Q.10. Find the molar specific heat (in terms of R) of a diatomic gas while undergoing the following

process. dQ = dW21

dU21

Q.11. Three identical metallic rods are arranged as an equilateral

triangle ABC as shown in the figure. Rod AB and BC have same thermal conductivity k. Ends A & C are maintained at constant temperature 1000C and 500C. Find the temperature of junction B and conductivity of rod AC such that heat flowing is AB is same as in AC. (provided that only conduction is takes place)

B

A C 1000 50

0

Q.12. An ideal gas undergoes a cyclic process ABCD as shown in the

figure. Draw the corresponding P-V diagram.

B

C

A

D

V

T Q.13. A vessel contains a mixture of one mole of CO2 and two moles of nitrogen at 300 k. Find the ratio

of the average rotational kinetic energy of CO2 molecules to that of N2 molecules. Q.14. An insulated container containing monoatomic gas of molar mass m is moving with velocity v0. If

the container is suddenly stopped, find the change in temperature. Q.15. Following graph shows variation of temperature with time of a

cooling body. The surrounding temperature is 270 C. Find tc.

1

2

500C

480C

400C

380C

3 4

tc t =5 min. time o

Temp. n

Q.16. Find the temperature of junction of three rods of the same

dimensions having thermal conductivity arranged as 3k, 2k, and k shown with their ends at 1000C, 500C and 00C?

1000C

00C

500C

T

3k

k

2k

Q.17. An insulated container containing monoatomic gas of molar mass m is moving with a velocity v0. If

the container is suddenly stopped, find the change in temperature. Q.18. An ideal gas, whose adiabatic exponent is equal to J is expended so that the amount of heat

transferred to the gas is equal to the decrease of internal energy, find (a) the molar heat capacity of the gas in this process.

http://www.rpmauryascienceblog.com/ (b) the equation of the process in the variables T, V. Q.19. A body cools from 620C to 500 in 10 min and 420 C in the next 10 minutes. Find the temperature of

the surrounding. Q.20. Two moles of Helium gases (J = 5/3) are initially at temperature 270 C and occupy a volume of 20

litres. The gas is first expanded at constant pressure until the volume doubles. Then it undergoes an adiabatic change until the temperature return to its initial value.

(a) Sketch the process on P-V diagram. (b) What are the final value and pressure of the gas. (c) What is the work done by the gas. Q.21. Taking the composition of air to be 75% Nitrogen and 25% oxygen by weight, calculate the velocity

of sound through air at STP. Q.22. A heater wire boils water in a given electric kettle in 3 minutes. Another heater wire boils water in

the same electric kettle in 7 minutes. Find the time taken when both heater wires are connected in series in the same electric kettle across the given mains.

Q.23. A circular hole of radius 2 cm is made in an iron plate at OoC. What will be its radius at 100oC? D

for iron = 11 u 10-6 /0C. Q.24. One mole of an ideal monoatomic gas is taken through the

thermodynamic process Ao Bo C as shown in figure. Find the total heat supplied to the gas.

3Po

Po

To 2To

C

B A P

T

Q.25. P-V diagram of a cyclic process is given in the figure. This is

elliptical process and O is the focus of the ellipse. Find the work

done in the diagram.

O

v0 8v0 V (m3)

N/m2

P0

4P0

Q.26. What is the heat input needed to raise the temperature of 2 mole of helium gas from 00C to 1000C

(take R = 8.31) (a) at constant volume ? (b) at constant pressure ? Q.27. A piece of ice of mass 100g and at temperature 00C is put in 200 g of water at 250C. Assuming that

the heat is exchanged only between the ice and the water, find the final temperature of the mixture. Latent heat of fusion of ice = 80 cal/g, specific heat capacity of water = 1 cal/g 0C.

Q.28. In steady state, the temperature at the end A and B of 20 cm long rod AB are 1000C and 00C. Find

the temperature of a point 9 cm from A.

http://www.rpmauryascienceblog.com/ Q.29. A cubical block of coefficient of linear expansion D floats in liquid of

coefficient of volume expansion J. When temperature of block and liquid is raised by t0C, it is found that immersed portion of cube in liquid remains same. Find the relation between J and D.

x0

Q.30. A hollow spherical ball of inner radius a and outer radius 2a is made of a uniform material of

constant thermal conductivity k. The temperature within the ball is maintained at 2T0 and outside the ball is T0. Find (a) the rate at which heat flows out of the ball in the steady state.

(b) the temperature at r = 2a3

Q.31. Two bodies A and B have thermal emissivity of 0.01 and 0.81 respectively. The outer surface areas

of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength OB corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 Pm. If the temperature of A is 5802 k, calculate

(a) the temperature of B (b) wavelength OB Q.32. The thickness of ice on a lake is 5.0 cm and the temperature of air is 200C. Calculate how long

will it take for the thickness of ice to be doubled. Thermal conductivity of ice = 0.005 cal/ cm sec. density of ice = 0.92 gm/cc. and latent heat of ice is 80 cal/g.

Q.33. A cubical tank of water of volume 2m3 is kept at a constant temperature of 650C by 2kW heater. At

time t = 0 the heater is switched off. Find the time taken by the tank to cool down to 500C given the temperature of the room is steady at 150C. density of water = 103kg/m3 and specific heat of water = 1 cal/gm-0C(Assume the tank to behave like a black body and cool according to Newtons law of cooling) Take 1 KW = 240 Cal/sec

Q.34. One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown. Calculate

P

3P0

P0 A

B

C

V0 V2V0

(i) Work done by the gas (ii) Heat rejected by the gas in the path CA & heat absorbed by the gas in the path AB. (iii) Net heat absorbed by the gas in path BC (iv) Maximum temperature attained by the gas during the cycle.

http://www.rpmauryascienceblog.com/ Q.35. One mole of an ideal monoatomic gas is taken along a cycle

ABCDA where AB and CD are adiabatic process. BC and DA are isochoric process. If vA/vB = 8, then find efficiency of the cycle.

B

C

D

A V

P

Q.36. A gaseous mixture has the following volumetric composition per mole of the mixture. He = 0.2, H2 = 0.1, O2 = 0.3, N2 = 0.4 Assuming the mixture to be a perfect gas, determine

(a) the apparent molecular weight of the mixture (b) cv and cp for the mixture (c) gas constant per kg of the mixture.

Q.37. At 270C two moles of an ideal monoatomic gas occupy a volume V. The gas expands a

adiabatically to to a volume 2V. Calculate (i) the final temperature of the gas.

(ii) change in its internal energy and (iii) the work done by the gas during this process.

Q.38. The speed of sound at N.T.P. in air is 332 m/sec. Calculate the speed of sound in hydrogen at (i)

N.T.P. (ii) 8190C temperature and 4 Atmospheric pressure. (Air is 16 times heavier than hydrogen)

Q.39. A cylinder with a piston holds a volume V1 = 1000 cm

3 of air at an initial pressure p1 = 1.1 u 105 Pa and temperature T1 = 300 K. Assume that air behaves like ideal gas. The sequence of changes imposed on the air in the cylinder is shown in the figure.

(a) AB the air heated to 375 K at constant pressure. Calculate the new volume, V2.

(b) BC the air is compressed isothermally to volume V1. Calculate the new pressure p2.

(c) CA the air cools at constant volume to pressure p1. Find the net work done on the air.

p2

p1

C

BA

v2v1

Q.40. A wall is made of 7.5 cm thick magnesia, surfaced with 0.5 cm thick steel plate on one side & 2.5

cm thick asbestos on the other side. The thermal conductivities of steel, magnesia and asbestos are 52.3, 0.075 & 0.081 W/m-K respectively. If the outer surface temperature of steel plate is 1500C and that of asbestos is 380C find

(a) Rate of heat loss per square meter of surface area of wall & (b) Interface temperatures. Q.41. A hollow spherical ball of inner radius a and outer radius 2a is

made of a uniform material of constant thermal conductivity K. The temperature within the ball is maintained at 2T0 and outside the ball it is T0.Find,

(i) the rate at which heat flows out of the ball in the steady state,

(ii) the temperature at r = 2a3

. Where r is radial distance from

a

2a

2T0

T0

K

http://www.rpmauryascienceblog.com/ the centre of shell. Assume steady state condition.

Q.42. A body cools down from 500C to 450C in5 minutes and to 400C in another 8 minutes. Find the

temperature of the surrounding. Q.43. At 270C two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically

to a volume 2V. Calculate (i) the final temperature of the gas.

(ii) change in its internal energy and

Q.44. A hot body of mass m, specific heat s is initially at temperature TI and obeys Newtons law of cooling when placed in a surrounding of temperature T0. Find

(a) the heat lost to the surroundings from t = 0 to the thermal equilibrium condition (b) the time taken for losing 60 % of the total heat lost. Q.45. The temperature of a body falls from 400C to 360C in 5 minutes when placed in a surrounding of

constant temperature 160 C. Find the time taken for the temperature of the body to fall from 360C to 320 C.

Q.46. A solid copper sphere (density U and specific heat C) of radius r at an initial temperature 200k is

suspended inside a chamber whose walls are at almost constant temperature at 0 K. Calculate the time required for the temperature of the sphere to drop to 100 K.

Q.47. Hot oil is circulated through an insulated container with a wooden lid

at the top whose conductivity K = 0.149 J/(m-qC-sec), thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at T" = 127q. If the ambient temperature Ta = 27qC. Calculate

(a) rate of heat loss per unit area due to radiation from the lid.

(b) temperature of the oil. (Given V = 3

17 u 108)

T"= 127qC

Ta = 27qC

Hot Oil

To

Q.48. An insulated vessel fitted with a frictionless movable piston of mass

m and area of cross section A consists of certain amount of mono-atomic gas. When the temperature of the gas was T0 the piston was at a height of 0 from the bottom of the vessel. Now the gas is heated such that piston raises to a height of 2"0 from the bottom then (assume that atmospheric pressure is P0)

(a) Identify the process? (b) work done by the gas ? (c) final temperature ? and (d) heat absorbed by the gas ?

A

T0

~

"0

P0

Piston m

Q.49. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of

mass M. The piston and the cylinder have equal cross-sectional area A. Atmospheric pressure is Po and the volume of the gas is Vo when the piston is in equilibrium. The piston is slightly displaced from the equilibrium position. Assuming the process to be adiabatic, show that the piston executes SHM. Find the angular frequency of oscillation.

http://www.rpmauryascienceblog.com/ Q.50. A copper and a tungsten plate having a thickness 2mm each are

riveted together so that at 00C they form a flat bimetallic plate. Find the radius of curvature of the layer common to copper and a tungsten plates at 2000C. The coefficients of linear expansion for copper and tungsten are 1.7 u 10-5 k-1 and 0.4 u 10-5 k-1 respectively.

copper

tungsten

Q.51. Three moles of an ideal gas (cp = 7/R) at pressure PA and temperature TA is isothermally expanded

to twice is initial volume. It is then compressed at constant pressure to its original volume. Finally the gas is compressed at constant volume to its original pressure PA.

(a) Sketch P V (P on X-axis, V on y-axis) and P T (P on X-axis, T on y-axis) diagrams for the complete process.

(b) Calculate the net work done by the gas and net heat supplied to the gas during the complete process. (ln2 = 0.693).

Q.52. A smooth vertical tube having two different sections is open from

both ends and equipped with two pistons of different areas as shown. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional S = 10cm area of the upper piston is 2 greater than that of the lower one. The combined mass of the two pistons is equal to m = 5.0 kg. The outside air pressure is po = 1.0 atm. By how many kelvins must the gas between the pistons be heated to shift the pistons through l = 5.0 cm.

[Take g = 9.8 m/s2]

po

po

Q.53. One mole of a monatomic gas is carried along process ABCDEA as shown in the diagram. Find the net work done by gas and heat given in the process.

A

D E

CB

O 1 2 3 4

4

2

1

3

P(N

/m2 )

Q.54. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand

adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of initial pressure and

volume. Q.55. Two moles of an ideal monatomic gas, initially at pressure p1 and volume V1 undergo an adiabatic

compression until its volume is V2. Then the gas is given heat Q at constant volume V2. (a) Sketch the complete process on a p-V diagram. (b) Find the total work done by the gas, the total change in its internal energy and the final change

in its external energy and the final temperature of the gas. [Give your answers in terms of p, V1, V2, Q and R]

http://www.rpmauryascienceblog.com/ Q.56. A reversible heat engine carries 1 mole of an ideal

monoatomic gas around the cycle as shown in the figure. Process 1o 2 takes place at constant volume, process 2o 3 is adiabatic and process 3o 1 takes place at constant temperature. (a) If the pressure at point (1) is one atmosphere find the

pressures at points (2) and (3). (b) Compute the heat exchanged, the change in internal

energy and the work done for each of the three processes and for the cycle as a whole.

3

2

1

P n

V o 300k

600k

(c) Compute the efficiency of the cycle. (Give all answers in terms of R) (ln2 = 0.693) Q.57. Heat flows radially outwards through a spherical shell of radius R2 and R1 (R2 > R1) and the

temperature of inner and outer surfaces are T1 and T2 respectively. Find the radial distance from the centre of the shell at which the temperature is just halfway between T1 and T2.

Q.58. Determine the work done by an ideal gas during 1 o 4 o 3 o 2 o 1.

Given P1 = 105 Pa, P0 = 3 u 10

5 Pa, P3 = 4 u 105 Pa and

v2 - v1 = 10 litres.

v1

P1

P0

P3

P

V v2

1 2

3 4

Q.59. An ideal monoatomic gas is used to operate an engine. P-V diagram

of the cyclic used is shown in the figure. Find the efficiency of the cycle.

4

32

1

V2V0V0

p0

3p0

p

Q.60. 2 moles of an ideal monatomic gas undergoes through the following changes in a cyclic process,

(i) Isothermal expansion from a volume 0.04 m3 to 0.10 m3 at 870 C (ii) at constant volume, cooling to 270 C. (iii) Isothermal compression at 270 to 0.04 m3 (iv) at constant volume heating to original pressure volume and temperature Then,

(a) Draw P-V diagram of the complete cycle (b) Find the heat absorbed by the gas (c) Find the work done by the gas during the complete cycle (d) Find the efficiency of the cycle. (e) Find the change in internal energy of the gas during the complete cycle.

Q.61. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 k. At

time t = 0 the temperature of X is T0 = 400 k. If cools according to Newtons law of cooling. At time

http://www.rpmauryascienceblog.com/ t1, its temperature is found to be 350 k. At this time t1, the body X is connected to a large box Y at atmospheric temperature TA, through a conducting rod of length L, cross-sectional area A and thermal conductivity k. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1.

Q.62. One mole of an ideal monoatomic gas is taken along the cycle ABCDA

where AB is isochoric, BC is isobaric, CD is adiabatic and DA is

isothermal. Find the efficiency of the cycle. It is given that 4TT

A

C ,

16

1

v

v

D

A and ln 2 = 0.693. D

B C

A

V

P

Q.63. The insulated box shown in figure has an insulated partition which can

slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a monatomic ideal gas (J = 5/3) at a pressure Po, volume Vo and temperature To. The chamber on the left is slowly heated by an electric heater so that its gas, pushing the partition, expands until the final pressure in both the chambers becomes 243Po/32. Determine :

(i) The final volume and temperature of the gas in B. (ii) The heat given to the gas in A by the heater.

A B

Q.64. Find the molar specific heat (in terms of R) of a diatomic gas while undergoing each of the following

processes :

(i) For any infinitesimal part of the first process, dQ = dW21

dU21

(ii) For the second process : pV2T = constant. Q.65. One mole of a gas is isothermally expanded at 270 C till the volume is doubled. Then it is

adiabatically compressed to its original volume. Find the total work done. (J = 1.4 and R = 8.4 J/mol -k).

Q.66. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand

adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of initial pressure and

volume. Q.67. An electrically heating coil was placed in a calorimeter containing 360 gm of water at 100 C. The coil

consumes energy at the rate of 90 watt. The water equivalent of the calorimeter and the coil is 40 gm. Calculate what will be the temperature of the water after 10 minutes ?

Q.68. A double-pane window used for insulating a room thermally from outside, consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state the room-glass interface and the glass-outdoor interface are at constant temperature of 270C and 00C respectively. Calculate the rate of heat flow through the window pane.

http://www.rpmauryascienceblog.com/ Also find the temperatures of other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 Wm1 K1 respectively.

Q.69. Three rods of material x and three rods of material y are connected a shown in figure. All the rods

are of identical length and cross-sectional area. If the end A is maintained at 600 C and the junction E at 100 C, calculate the temperature of junctions B, C and D. The thermal conductivity of x is 0.92 cal/cms C0 and that of y is 0.46 cal/cm s C0.

x

x x

y y

E

C

D

A y

600C 100C

(A)

B

Q.70. A thermally insulated vessel is divided into two parts by a heat

insulating piston which can move in the vessel without friction. The left part of the vessel contains one mole of an ideal mono-atomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel. Determine the heat capacity C of the system, neglecting the heat capacities of the vessel, piston, and spring.

Q.71. When a block of iron floats in mercury at 00C, a fraction K1 of its volume is submerged, while at

temperature 800C fraction K2 is seen to be submerged. If the coefficient of volume expansion of iron is JFe and that of mercury is JHg. Find the ratio K1 / K2.

Q.72. Two rods of different metals having the same area of cross-section A, are placed end to end

between two massive walls. The first rod has a length "1, co-efficient of linear expansion D1 and Youngs modulus of elasticity Y1. The corresponding quantities for second rod are "2, D2 and Y2 respectively. The temperature of both the rods is now raised by T degrees. Find the force which the rods exert on each other at the higher temperature in terms of the given quantities. Assume, there is no change in the area of cross-section of the rods, the rods do not bend and there is no deformation of the walls.

Q.73. Two moles of an ideal monatomic gas is taken through a cycle

ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of the gas vary such that PT = constant. If T1 = 300 K, calculate

(a) The work done on the gas in the process AB (b) The heat absorbed or released by the gas in each of the

processes.

2P1

P1

B C

A

T1 2T1 T

P

Give answers in terms of the gas constant R.

http://www.rpmauryascienceblog.com/ Q.74. Two moles of a certain ideal gas at temperature T0 = 300K were cooled at constant volume so that

the gas pressure reduced = 2 times. Then as a result of constant pressure process, gas expanded till its temperature got back to initial value. Find the total amount of heat absorbed by the gas in this process.

Q.75. Two identical sphere with surface area A and emissivity e are connected by a metal rod of length ",

with high conductivity k and area of crossection a (a T2 > T0 . Temperature difference between sphere and surrounding is small enough to consider Newtons laws of cooling for heat loss through radiation. Find the temperature difference between spheres as function of time.

Q.76. A polyatomic gas is initially taken at a pressure P0 and temperature T0 and volume

V1 = 500 ml. When the gas is compressed adiabatically to volume V2 = 100 ml the temperature increases from T0 to T1 and the pressure from P0 to P1 . Now the pressure increases from P1 to P2 isochorically and the temperature rises to T2 . The gas is expanded adiabatically from volume V2 to V1 such that the temperature drops to T3 . Finally the gas pressure drops to P0 isochorically . Find efficiency of this cycle if J = 1.33.

Q.77. One mole of a mono atomic ideal gas is taken through the cycle

shown in figure. A o B Adiabatic expansion B o C Cooling at constant volume C o D Adiabatic compression. D o A Heating at constant volume The pressure and temperature at A, B etc,. are denoted by PA,

TA; PB, TB etc. respectively. V

P DC

B

A

Given TA = 1000K, PB = (2/3)PA and PC = (1/3)PA. Calculate (a) The work done by the gas in the

process A o B (b) the heat lost by the gas in the process B o C and (c) temperature TD. Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.

Q.78. A fixed mass of gas is taken through a process A o B o Co A. Here A o B is isobaric. B o C is

adiabatic and C o A is isothermal. Find (a) Pressure and volume at C (b) work done in the process (take J = 1.5)

A B

C

1 4

10 5

P (N/m2)

V(m3) Q.79. One mole of an ideal gas whose pressure changes with volume as P = Dv where D is a constant, is

expanded so that its volume becomes K times the original. Find the change in internal energy and heat capacity of the gas.

http://www.rpmauryascienceblog.com/ Q.80. Consider the shown diagram where the two chambers separated by

piston-spring arrangement contain equal amounts of certain ideal gas. Initially when the temperatures of the gas in both the chambers are kept at 300 K. The compression in the spring is 1 m. The temperature of the left and the right chambers are now raised to 400

K and 500 K respectively. If the pistons are free to slide, find the final compression in the spring.

300 K 300 K

1 m 1 m

Vaccume300 K

300 K

Q.81. One mole of a gas is taken in a cylinder with a movable piston. A

resistor R connected to a capacitor through a key is immersed in the gas. Initial potential difference across the plates of the capacitor is

equal to 213

13 V. When the key is closed for (2.5 ln 4 ) minutes , the

gas expands isobarically and its temperature changes by 22 K . (a) Find the work done by the gas (b) Increment in the internal energy of the gas (c) The value of J

C = 75mF

R = 2k:

Q.82. One mole of a monoatomic ideal gas has been subjected to an

isochoric-isobaric cycle ABCDA. The temperature at D is T0 = 200 K. Calculate (i) The temperatures at the points A, B, C of the cycle. (ii) The heat absorbed / released in each of the processes AB, BC,

CD, DA. (iii) The net work done in the process and the efficiency of the cycle.

C

BA

D

V3V0V0

p0

2p0

p

Q.83. Find the number of strokes that the piston of an air pump must make in order to pump a vessel of

volume Vc.c from a pressure P1 to P2 if the change in volume corresponding to one stroke is v1cc. Assume that the air in the vessel is in good thermal contact with the surroundings.

Q.84. A pendulum clock consists of an iron rod connected to a small heavy bob. If it is designed to keep

correct time at 200C, how fast or slow will it go in 24 hours at 400C. (Diron = 12 u 10

-6 /0C) Q.85. A vessel of volume 2000 cm3 contains 0.1 mole of oxygen and 0.2 mole of carbon dioxide. If the

temperature of the mixture is 300 K, find its pressure. Q.86. At the top of a mountain a thermometer reads 70C and a barometer reads 70 cm of Hg. At the

bottom of the mountain they read 270C and 76 cm of Hg respectively. Compare the density of the air at the top with that at the bottom.

Q.87. A body of mass 25kg is dragged on a horizontal rough road with a constant speed of 20km/hr. It

the coefficient of friction is 0.5, find the heat generated in one hr. if 50% of the heat is absorbed by the body, find the rise in temperature. Specific heat of material of body is 0.1Cal/gm-0C. (4.2 J = 1 Cal., g = 10m/s2)

Q.88. One mole of an ideal gas whose pressure changes with volume as P = Dv where D is a constant, is

expanded so that its volume becomes K times the original. Find the heat supplied to the gas.

http://www.rpmauryascienceblog.com/ Q.89. A hot body placed in air cooled according to Newtons law of cooling, the rate of decrease of

temperature being k times the temperature difference from the surrounding, starting from t = 0, find the time in which the body will loose 90 % of the maximum heat it can loose.

Q.90. 1 mole of diatomic gas is kept in cylinder of volume V0 at

temperature T0 and pressure P0. Now gas is taken through three processes as shown in figure. Process 1 o 2 follows law P = V, Process 2o 3 is adiabatic while process 3o 1 is isothermal. Find (a) temperature at state 2 in terms of T0

(b) efficiency of cycle.

2v0 v0

P0

2P0

P

1 3

2

SOLUTION

Q.1. dQ 1 (20 /100)dt k A

= (100- 0) (i)

T 100(A

)100/9(

k

1

dt

dQ (ii)

from (i) and (ii), T = 550C

Q.2. 21 f

nmv nR T2 2

(i)

f = 1

2J

(ii)

from (i) and (ii)

T = )1(R2

mv2J

Q.3. From B to A, 0 = UBA + WBA UBA = + 30 From A to B 20 = UAB + WAB 20 = -30 + WAB WAB = 50 Q.4. T = T2 T1 = 78.3 40.6 = 37.7qC T = ( T1 + T2)

http://www.rpmauryascienceblog.com/ = r (0.2 + 0.3) = r 0.5 qC = (37.7 r0.5)qC. Q.5. WAB = Area under A o B bounded by volume axis.

= 10 u 103 (25 10) u 10-6 + 21

(25 10) u 10-6 u 20 u 103

= 0.15 + 0.15 = 3.0 J. In path A o C o B = 30 u 103 (25 10) u 10-6 = 0.45 J.

Q.6. dQ 1 (20 /100)dt k A

= (100- 0) (i)

T 100(A

)100/9(k1

dtdQ

(ii)

from (i) and (ii), T = 550C Q.7. P2V = constant

also PV

cons tan tT

2

2

TV cons tan t

V

? T2V-1 = constant

? 2 2 0f 00

3VT T

V

' Tf = 3 T0.

Q.8. WAB = 0

QAB = vAB + WAB 600 = UAB QBC = UBC + WBC 200 = UBC + 8 u 10

4 u 3 u 10-3 UBC = - 40 UAC = 560

P

A

B C

V

PA

PB

VA VC

Q.9. dQ dU dWndt ndT ndT

C = Cv +

'

ndTPdv

P = Dv2 PV = nRT PdV + VdP = nRdT also from P = Dv2 dP = 2DV dV PdV + 2DV2 dV = nRdT 3P dV = nRdT

3R

ndTPdV

? C = R611

3R

R23

http://www.rpmauryascienceblog.com/

Q.10. Process: dQ = - dW21

dU21

Ist law: dQ = dU + dW

? dU + dW = - dW2

1dU

2

1 dW = - 3 dU

dQ = dU - 3dU = -2dU

? C = R5C2ndTdU

2ndTdQ

V

Q.11. Let " and a be the length and cross-section area of each rod.

? QAB = QBC ? ""

)50T(ka)T100(kA

? T = 750C

also if QAB = QAC ? ""

)50100(ak)75100(ka c

? 25 k = 50 k ? kc = k/2 Q.12.

D A

C B

V o

P n

Q.13. Average rotational K.E. = 2

nRTfr

where fr = rotational degree of freedom )

Avg. rotational KE of CO2 = RT2

2TR1

uuu

Avg. rotational KE of N2 = RT22

2TR2

uuu

Ratio = 2

1.

Q.14. Applying COE

TnCmv2

1v

20 = TR2

3

M

m

? T = R3

Mv20

Q.15. Using Newtons law of cooling

)(kt gsurroundinavg

TT

T

for 1, 2


?

'

27

2

4050k

5

4850

)2749(k5

2 (i)

for 3, 4

'

c

272

3840k

t3840

)2739(kt

2

c (ii)

on solving (i) and (ii) we get tc = 9.1 min.

Q.16. R1 = kA3"

, R2 = kA2"

R3 = kA"

100 - T = I1R1 , T - 50 = R2I2, T - 0 R3 (I1 I2)

T = 3

200 0C

Q.17. Loss in K.E. of the gas E = 21 (nm) 20v , where n = number of moles.

If its temperature change by T.

Then n23 R T =

21 (nm) 20v

T = R3

mv20 .

Q.18. (a) Q = u1 U2 = -(u2 u1) = - u

dQ = - du = -n CvdT = -1

nRJ

dt

C = 1

RdT

dQJ

Q

(b) not av ailable Q.19. For first ten minutes

10

5062

dt

dT = - 1.2 0C/min

T = 0T105062

= (56 T0)

0C

-kA(56 T0)0 = -1.2 0C / min. (i)

Similarly for next 10 minutes

10

5042

dt

dT 00 = - 0.8 0C/min


T =

'

2

5042 - T0 = (46 T0)

0C

-0.80C / min = -kA (46 T0)0C (ii)

dividing (i) and (ii) T0 = 26

0 Q.20. (a) For an ideal gas

P = V

nRT =

51020

3003.82

u

uuN/m2

= 2.5 u 105 N/m2

(b) TA = T, TB = 2T at B, PB = PA =2.49 u 10

5 N/m2

Pn

V o

A

C

B

VA VC VB

vB = 2vA = 40 u 10

-3 m3, TB = 600 k from TVr-1 = constant at B and C

1

B

c 2v

v J = 23/2

vc = 2 2 vB = 2 u 1.414 u 40 = 113 "

Pc = 31013.113

3003.82

VC

NRTCu

uu

= 0.44 u 105 N/m2

(c) WAB = 2.49 u 10

5 (40 20) u 10-3 = 4980 J

W2 = )3/5(1

3.82]TT[

1rnR

12 u

[300 600]

= 7470J Wnet = 4980 + 7470 = 12450 J

Q.21. Molecular weight of the mixture is given by )M/m(Mm

6 6

? M =

32

25

28

752575

= 28.9

J of the mixture given by

1

n1

n1

nn

2

2

1

1

m

21

J

J

J

Jm = 1.4

? velocity of sound v = M

RTJ = 331.3 m/s

Q.22. 2v

H .tR

2v

R .t k.tH

1 1

2 2

R tR t


eq eq

1 1

R t

R t

1 2 2eq 1 1 1 21 1

R R tt .t 1 t t t 10 mins

R t

Q.23. R R T100 0 1 ( )D

> @ ( ) ( / )( )2 1 11 10 1006cm x C Co o ( )( )2 1 11 10 4cm x

= 2.0022 cm. Q.24. A o B represents an isobaric process,

? QAB = 1 u 0 0 05 5

R(2T T ) RT2 2

B o C represents an isothermal expansion ? UBC = 0

QBC = 1.R.2T0ln 00

3PP

'

= 2RT0ln3

? Q = RT0[2.5 + 2ln3] Q.25. W = area enclosed by the ellipse = Sab = S (3P0) (7v0) Nm = 21SP0v0 Nm

O

v0 8v0 V (m3)

N/m2

a b 4P0

P0

Q.26. (a) At constant volume Q = nCv T = 2 (3R/2)100 = 300 R = 300 u 8.31= 2493 Joules (b) At constant pressure Q = nCp T = 2 (5R/2) (100) = 500 u 8.31 = 4155 Joules Q.27. The heat required to melt the ice at 00C = 100 u 80 = 8000 cal. The heat given by water when it cools down from 250C to 00C = 200 u 1 u 25 = 5000 cal. Clearly, the whole of the ice can not be melted, as the required amount of heat is not provided by

the water. Therefore, the final temperature of the mixture is 00C.

Q.28. dQ 1 (20 /100)dt k A

= (100- 0) (i)

T 100(A

)100/9(k1

dtdQ

(ii)

from (i) and (ii), T = 550C Q.29. Let A0 be the cross-section of cube and U0 the density of liquid before temperature rise.

http://www.rpmauryascienceblog.com/ After t0C increase in temperature, the density of liquid becomes

U = 0(1 t)

U

J

while new cross-sectional area of cube is, A = A0 (1 + 2D t) Since Mg = A0x0 U0g . . . (i) where x0 is the length of cube in liquid and M is the mass of cube. Also Mg = Ax0 Ug . . . (ii)

A0 x0U0g = A0(1 + 2D t) 0 0x

g(1 t)

U

J

J = 2D.

Q.30. dTr4

drk1

dtdQ

2

S

'

S

12

CdT 1dr 4 k r

Integrating,

T = 21 C

r1

k4C

S

At r = a, T = 2T0 and at r = 2a, T = T0

C2 = 0, C1 = 8SakT0 ? T = 0Tra2

(i) 0akT8dtdQ

S

(ii) T(r = 04T3a

)2 3

Q.31. (a) eA VAA 4BBB

4A TAT VU

0.01 u (5802)2 = 0.81 (TB)

4 TB = 1934 k. (b) OA TA = OBTB OB - OA = 1 Pm OB = 1.5 Pm.

Q.32.

'

dtdm

LdtdQ

, LAy

)]20(0[kA

U

dt

dy

U t

0

10

5

dtL

k20ydy , T =

2 210 52 2

'

t = 27600 sec. = 7 hrs. 40 minutes. Q.33. We have heat supplied by heater = heat lost by tank by radiation under steady state. ? 2 = k (65 15) where k is a constant ? K = 2/50 = 2(240)/50 = 48/5 Cal/s-0C At any instant if the temperature of the tank be T then

we have )T(s.m

KdtdT

15


or dt.s.m

KT

dT

15

or - tdt

s.mK

TdT

0

50

65 15

-ln 5065

15T = s.m

K.t

ln

15501565

= s.m

K.t

or t =

'

'

7

10

548

102103550 33

lnxx

lnK

s.m

= 20 hrs. (Approximately) Q.34. (i) work done = area under the curve.

= > @> @0000 3221

PPvv = P0v0

(ii) heat rejected in the path CA = nCP.dT (constant pressure process)

= 1 x AC TTR 25

=

252

25 000000 vP

RvP

RvPR

heat absorbed in path AB = nCV.dT(constant vol process.

= > @

RvP

RvPR

TTRxx AB00003

23

23

1

= 3 P0v0 (iii) for ABC, Q = W (cyclic process)

? 000000 325

vPQvPvP BC

? QBC = P0v0 3P0v0 + 0025

vP

= 2

00vP

(iv) T

Pv= constant so, when Pv is maximum, T also is maximum.

Pv is maximum for process BC. Hence temperature will be maximum between B & C. Let equation for BC be P = Kv + K1 satisfying both points B and C For B, 3P0 = Kv0 + K1 For C, P0 = K (2v0) + K1

So K = 0

02

v

Pand K1 = 5P0

? equation for BC becomes P = 00

0 52

Pvv

P

or 00

00 52

Pv

vP

v

RT or T =

0

20 25

vv

vRP


for maximum 0 dvdT

? 04

50

0

vv

RP

or v = 4

5 0v

? Tmax =

'

'

0

2000 1

45

24

55

vvv

RP

=R

vP8

25 00

Q.35. K = pliedsup

net

)Q(W

K = BC

DABC

QQQ

= 1 - BC

AD

TTTT

Let kv

v

v

v

C

P

B

A

1

C

D

B

A

k

1

T

T

T

TJ

'

K = 1 - 1)k(

1J

= 1 - 3/2)8(

1 = 75 %.

Q.36. Table of determination

Constituent Molar fraction

Molecular weight

Mass/mole of mixtur

e

Cv Specific heat per

mole of mixture

He 0.2 4 0.8

2

3R

0.3 R

H2 0.1 2 0.2

25

R 0.25 R

O2 0.3 32 9.6

25

R 0.75 R

N2 0.4 28 11.2 2

5R

R

(a) For this mixture mass per mole = 21.8 (b) Specific heat at constant volume for the mixture = (0.3+0.25+0.75+1)R = 2.3 R = 19.09 J mole-1k-1 and cp = cv + R = 27.39 J mol

-1 k-1

(c) gas constant per kg = MR

= 8.21

3.8 = 0.4 Jkg-1 k-1

http://www.rpmauryascienceblog.com/ Q.37. (i) TI

1yiV

= Tf 1y

fV T2 = 189 k

(ii) U = nCv T

where Cv =2

3R and T = -111 k

V = 2 u 23

u (-111) = -2767.2 (J)

(iii) In adiabatic proces U = - W W = 2767.2 J

Q.38. (i) we have velocity of sound in a medium given by U

JP

at N.T.P. U

U a

a

H

v

v

at N.T.P. H

a

a

H

v

v

U

U = (16)1/2 = 4

? vH = 4va = 4 (332) = 1328 m/s.

(ii) we also have v v T

? 273

819273

v

v

0

819 = 2

? v819 =2v0 = 2 (1328) = 2656 m/s

Q.39. (a) V2 = 31

21 cm1250300

3751000

TT

V

(b) p2 = 1

21

VVp

=

Pa10375.11000

1250101.1 55

u u

(c) Wnet = WAB + WBC + WCA WAB = p1(V2 - V1) = 1.1 u 10

5 (1250 - 1000) u 10-6 = 27.5 J

WBC = p2V1 ln 2

1

VV

= (1.375 u 105)(1000 u 10-6) ln 1250

1000 = -30.7 J

WCA = 0 Wnet = 27.5 - 30.7 = -3.2 J, Work done on the gas = 3.2 J

Q.40.

3

3

2

2

1

1

41

KL

KL

KL

TTdtdQ

q = > @ 2100810520750573525038150

..

..

..

= 85.58 w/m2

T1 T2 T3 T4

K3K2K1

2.5 cm7.5 cm0.5 cm

and T2 = T1 q

'

1

1

KL

=150 85.85 21035250

'

..

http://www.rpmauryascienceblog.com/ = 149.990C

T3 = T4 + q

'

3

3

KL

= 64.40C

Q.41. In the steady state, the net outward thermal current is constant, and does not depend on the radial

position.

Thermal current, C1 = dr

dT)r4.(K

dt

dQ 2S

'

2

1

r

1

K4

C

dr

dT

S

Integrating, T = 21 C

r

1

K4

C

S

At r = a, T = 2T0 and at r = 2a, T = T0

C2 = 0, C1 = 8SaKT0 ? T = 0Tra2

(i) dt

dQ = 8SaKT0 (ii) T(r = 3a/2) = 4T0/3

Q.42. > @05.47k54550

T

Where T0 is the temperature of the surrounding.

> @05.42k84045

T

]5.42[k

]5.47[k

5

8

0

0

T

T

1.6 [ 42.5 - T0 ] = 47.5 - T0

68 - 1.6 T0 = 47.5 - T0

0.6 T0 = 20.5

T0 = 340.

Q.43. (i) Ti 1iV

J = Tf 1fVJ T2 = 189 k

(ii) U = nCv T

where Cv =23

R and T = -111 k

U = 2 u 2

3R u (-111) = -2767.2 (J)

Q.44. Under equilibrium condition temperature of the body = T0 Heat lost = Decrease in internal energy = ms (Ti - T0)

Newtons law of cooling )TT(kdtdT

0 where k is a constant


? ln ktTT

TT

0i

0

when 60 % of the total heat is lost ms (TI - T) = 0.6 ms (Ti - T0) substituting for T

t = k1

ln 2.5.

Q.45. )(kdtd

0TT T

TTT

36

40 0

kd

(5 min.)

k = - .min5)6/5ln(

k = TTT

32

36 0

ktd

t = )6/5ln()5/4ln(

u 5 min.

Q.46. V 4Sr2T4 =

'

'

US

dtdT

Cr34 3

dt = - 4T

dT3rcV

U

t = - s1072

rc7

T

dT3rc 6

100

2004

uV

U

VU

Q.47. (a) dtdQ = VHA[(T")

4 (Ta)4],

Rate of heat loss per unit area = 595 watt / m2. (b) Let To be the temperature of the hot oil

A595t

TTKA o "

To | 420 K or 147 0C

Q.48. (a) Isobaric process

(b) W = P(v2 v1) = (P0 + A

mg) "0A

(c) 2

2

1

1

T

v

T

v

2

0

0

0

T

A2

T

A "" T2 = 2T0

(d) Q = nCp T = 0

00

RT

AA

mgP "

'

2

RT5 0 = 00 AA

Mg

2

5"

'

U

http://www.rpmauryascienceblog.com/ Q.49. For adiabatic process PVJ = constant. In the equilibrium state, total pressure,

P = Po + A

Mg, and initial volume = Vo

Thus P JoV = (P + P)(Vo + V)J = (P + P) JoV

'

J

oVV

1

? P = J PoVV

Restoring force F = PuA = J PA oVV

= xA

MgP

V

Ao2

o

2

'

J

? Acceleration a = xA

MgP

MVA

mF

oo

2

'

J

Hence the motion is SHM with angular frequency Z = oo2 MVx

AMg

PA

'

J

Q.50. "C = "0 (1 + DC T) "t = "0 (1 + DT T)

T1T1

)2/dR(2/dR

T

C

DD

TT

R = 0.77 m

T

R

Q.51. (a) Using PV = nRT

V

2vA

PA/2

C A

B

vA

PA P

TA/2

PA PA/2 P

TA

C

B A T

(b) WAB = PdV = AT26.17dvvnRT

WBC = PdV = AAAA T45.12)v2v(2

P

WCA = 0 Net work done = 17.26 TA 12.45 TA = 4.81 TA As initial and final states of the gas are same U = 0 Q = U + W Q = W.


Q.52. T = nR

PV

V1 = A1X1 + A2X2 V2 = A1(X1 +") + A2 (X2 - ") = V1 + "(A1 - A2) = V1 + "s Net upward force due to inner and outer pressure (P1 - P0) A1 - (P1 - P0)A2 = mg (P1 - P0) s = mg

A1

A2

X1

X2

P1 =s

mg

+ P0 = P2 (Pressure remains same for equilibrium)

T = T2 - T1 = nR

VPVP 1122 = nR

sP

s

mg0

"

'

= nR

sPmg 0"" = 0.91 K.

Q.53. Process ABCA is clockwise while process ADEA is anticlockwise in P-V diagram Net work done

Area ABCA - area ADEA = 21

1221

uu u 1 u 1 = 21

(J)

During process ABCDEA W = 1/2 J U = 0 (cyclic process) Q = U + W = 1/2 J.

Q.54. (a) 122

111 VTVT

JJ

? TVJ-1 = 2

T (5.66 V)J-1

J - 1 = 7334.1693.0

66.5ln2ln

= 0.4

J = 1.4

? J = 1 + F

2 F = 5

(b) WA = J

J

1VPVP

1)TT(nR iiFFiF

from PV = nRT PF VF = PR2

T =

2

PV

WA =

2PV

PV4.0

1 = 1.25 PV.

http://www.rpmauryascienceblog.com/ Q.55. (a) A o B adiabatic compression B o C Heating at constant volume

(b) WAB = - )VPVP(1

nR2211 J

n = 2, J = 5/3, P2 = P1 3/5

2

1

V

V

'

? WAB = - > @23/521111 V)V/V(PVP1)3/5(R2

'

3/2

2

111 V

V1VP

2R3

C

Pn

P2

P1

V2 V1 oV

U = UAB + UBC

= Q - 2R3

P1V1

'

3/2

2

1

V

V1

For BC Q = nCv T

T =

2

R32

Q

u =

R3Q

For point A : P1V1 = 2RTA

For Point B : P2V2 = 2RTB .

For adiabatic change

P1 JJ 221 VPV P2 = P1 J

'

2

1

V

V

Further TB = R2VP

nRVP 2222

= R2

V

'

J

2

11 V

VP

? Final temperature TC = TB + T

= R3

Q

V

VP

R2

V3/5

2

11

2

'

=

R3Q

R2VVP 3/22

3/511 .

Q.56. (a) 1o 2 isochoric process 2

2

1

1

TP

TP

1 u 300600

= P2 = 2 atm

2o 3 adiabatic process P1-J TJ = constant


P3 = )3/5(1

3/5

2

1/

3

2

300

600P

T

T JJ

'

'

2 atm.

P3 = 2-3/2 atm =

22

1.

(b) 1o 2 (isochoric process) 2o 3 (adiabatic process) 3o 1 (isothermal process) W12 = 0 Q12 = n Cv T

= 1r

nR

(T2 - T1)

1)3/5(R

(600 - 300)

= 450R units. Q = U + W Q12 = U12 = 450 R units

W23 = 1

nRJ

(T1 - T2)

= 13/5

)300600(R1u

= 450 R units Q = 0 U = - W U = - 450 R units

W31 = nRT ln (P3/P1) = 1 u R u 300 ln 2-3/2 | -312 R units U = 0 Q = W Q31 = -312 R units.

Hence total work done in the cycle W = W12 + W23 + W31 = 138 R

(c) Efficiency of the cycle

K = absorbedHeat

doneworktotal =

R450R138

= 450138

| 0.31

Q.57. Rate of heat flow through a concentric shell of radius x and thickness dx is

Q = k4S x2dxdT

Or T

'

S d

Qk4

x

dx2

Integrating

T

TT

'

S

2

1

2R

1R 2d

Qk4

x

dx

or Q =

12

2121

RR

RRk4

TTS . . . (1)

Integrating equation (1) from R1 to r

Q

k4

rR

Rr 1

1

1 TTS

Q =

1

11

Rr

rRk4

TTS . . . (2)

Equating (1) and (2) and substituting T = (T1+T2)/2, We get r = (2R1R2)/(R1+R2) Q.58. From the figure

5 5

4 35 5

v v 4 10 3 1010 3 10 10

u u

u

v4 v3 = 5 litres


Now work done, W = 355 1010521

1021021 u

'

uuuuuu = 750 J.

Q.59. Process 1 - 2

1W2 = 0, 1Q2 = U2 - U1 = nCv(T2 - T1)

= n )PP(nR

v

2

R312

0

'

'

= 3P0v0 Process 2 - 3

2W3 = 3P0(2v0 - v0) = 3P0v0

2Q3 = nCp (T3 - T2) = n )VV(nR

P3

2

R523

0

'

'

= 00vP2

15

Process 3 - 4 3W4 = 0, 3Q4 = nCv (T4 - T3) < 0 as T4 < T3 Process 4 - 1 4W1 = P0 (v0 - 2v0) = - P0v0

4Q1 = nCp (T1 - T4) < 0 as T1 < T4

? efficiency = addedHeatworkNet

=

0000

00

vP32vP

15

vP2

= 04.1921

4 %.

Q.60. x (a) The P-V diagram will be as shown, x For isothermal expansion A o B

Q1 = nRT ln

'

1

2

v

v

= 2 u 8.314 u 360ln

'

04.010.0

= 5483.25J For isochoric process B o C

A

B

D

C

P

V

Q2 = nCvT = nCv (Tc TB) = 2 u 23

R u ( 300 360)

= -3 u 8.314 u 60 = -1496.5 J For isothermal compression C o D

Q3 = nRTln

'

C

D

v

v = 2 u 8.314 u 300 ln

'

10.004.0

= -4569.4J For isochoric process D o A

Q4 = nCvT = 2 u 23

u 8.314 u (360 300)

= 1496.5 J (b) Heat absorbed by the gas during the cycle

http://www.rpmauryascienceblog.com/ = 5483.25 + 1496.5 = 6979.75J (c) x Work done by the gas during the cycle

= Q1 + Q2 + Q3 + Q4

= 5483.25 - 1496.5 4569.4 + 1496.5

= 913.85 J

(d) x Efficiency of the cycle = supplied Heat

done workNet

= 75.69791.913

= 0.131.

(e) x Change in internal energy of the gas during the complete cycle = 0

Q.61. For cooling, dt

dms

dt

dQ T = -C

dt

dT ( heat capacity ms = C)

and from Newtons law of cooling

)(adt

d0TT

T

? -C )300(adt

dT

T (a constant , T0 - temperature of surrounding)

? TT

T t

0400

dtC

a

)300(

d

ln)300400(

)300(

T = -

Ca

.t

T = 300 + 100 t.

c

a

e

At time t = t1 ; T = 350 Hence, a = C ln (2/t1) Now, when the body X is connected to body Y

conductiondt

d

dt

d

'

T

T +

radiationdt

dQ

'

-C )(aL

)(kA

dt

d0

0 TTTT

T

TT

TCQ

LCkA

)(dtd

0

T

TT

TF

350 0)(

d = -

1t

2ln

LC

kA

1

!

t3

tdt

or, ln

T

1

F

t

2ln

LC

kA

300350

3002t1

? TF = 300 + 50 exp

'

11 t

2ln

LC

kAt2

Q.62. Taking the temperature, pressure and volume at D to be T0 , P0 and V0 using the relations. TA = TD, PAvA = PDvB for path DA

Phy 1 (11)

Documents

Transcript of Phy 1 (11)