Q1.

(1)

Heat transfer mechanisms: conduction; convection and radiation

Example of simultaneous heat transfer: the heating of water by an immerged heater is

through heat conduction and convection.

(2)

Thermal conductivity:

carbon nanotube >graphite > copper > bronze > window glass.

(3)

Based on the Newton’s law of cooling

)( airTThAQ eggs −=&

For natural heat convection, h is between 2-25 W/m2 °C. For force heat convection, h is

between 25-250 W/m2 °C. So the rate of heat convection is enhanced by using a fan for

the cooling of the boiled egg.

Q2.

Q3.

(1)

The problem is a steady state heat transfer process and can be summarized as follows

The area of the windows and the thermal resistances are

The steady rate of heat transfer through window glass then becomes

C/W 2539.0

0167.01923.0)0016.0(20417.02

C/W 0167.0)m 4.2(C).W/m 25(

11

C/W 1923.0)m 4.2(C)W/m. 026.0(

m 012.0

C/W 0016.0)m 4.2(C)W/m. 78.0(

m 003.0

C/W 0417.0)m 4.2(C).W/m 10(

11

2,211,

o

2o2

2

2,o

2

2

22

2

1

1glass31

22

1

1,i

°=

+++=+++=

====

°=°

===

°=°

====

°=°

===

convconvtotal

conv

air

conv

RRRRR

AhRR

Ak

LRR

Ak

LRRR

AhRR

2m 4.2m) 2(m) 2.1( =×=A

W114=°

°−−=

−=

∞∞

C/W2539.0

C)]5(24[21

totalR

TTQ&

(2)

The inner surface temperature of the window glass can be determined from

(3)

Heat cannot be conducted through an evacuated space since the thermal conductivity of

vacuum is zero (no medium to conduct heat) and thus its thermal resistance is infinitely

large. Therefore, if radiation is disregarded, the heat transfer through the window will be

zero. Then the answer of this problem is zero since the problem states to disregard

radiation. And the temperature of inner window glass is 24 degree C.

C19.2°°−=−=→−

=∞

∞ =C/W)W)(0.0417 114(C24o

1,11

1,

11conv

conv

RQTTR

TTQ &&