PROBLEM 1.47
KNOWN: Dimensions of a milk carton. Temperatures of milk carton and surrounding air.
Convection heat transfer coefficient and surface emissivity.
FIND: Heat transferred to milk carton for durations of 10, 60, and 300 s.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat transfer from bottom surface of milk carton and from top
surface since it is not in contact with cold milk, (2) Radiation is to large surroundings at the air
temperature.
ANALYSIS: The area of the four sides is A = 4L × w = 4(0.2 m × 0.1 m) = 0.08 m2. Thus,
()
44
conv rad sur
22 824244
()()()
10 W / m K 0.08 m (30 C 5 C) 0.90 5.67 10 W/m K 0.08 m (303 K) (278 K)
20.0 W 10.0 W 30.0 W
ss
qq q hATT AT T
εσ
∞
−
=+= −+ −
=⋅× °−°+×× ⋅× −
=+=
For a duration of 10 s,
Q = q∆t = 30.0 W × 10 s = 300 J <
Similarly, Q = 1800 J and 9000 J for durations of 60 and 300 s, respectively. <
COMMENTS: (1) The predicted heat transfer rates do not account for the fact that the milk
temperature increases with time. If the increase in milk temperature were accounted for, the values of
Q would be less than calculated. (2) If the coefficient of performance of the refrigerator is 2, COP =
Q/W = 2, then the required work input to re-cool the milk after leaving it in the kitchen for 300 s is
4500 J. At an electricity price of $0.18/kW⋅h, this would cost about $0.0002, which is insignificant.
Preventing bacterial growth is a more important reason to return the milk to the refrigerator promptly.
(3) The analysis neglects condensation that might occur on the outside of the milk carton.
Condensation would increase the rate of heat transfer to the milk significantly, increasing the
importance of returning the milk to the refrigerator promptly.
L = 200 mm
w = 100 mm
T
m
= 5°C
ε
= 0.90
T
∞
= 30°C
h= 10 W/m
2
·K
Air
T
sur
Milk
q
″
conv
q
″
rad