Polystyrene foam icebox has a total area of 0.950m2 and walls with a thickness of 28 mm. The box contains ice, water, and canned beverages at 0oC. The inside of the box is kept
cold by melting ice. 0.010 o W kC m
.
How much ice melts in one day if the icebox is kept in the trunk of a car at 38oC? Hint: Energy = Q = P.t
Answers
Answer:
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.
Solution
Identify the knowns.
A
=
0.950
m
2
;
d
=
2.50
cm
=
0.0250
m
;
T
1
=
0
∘
C
;
T
2
=
35.0
∘
C
;
t
=
1
day
=
24
hours
=
86
,
400
s
.
A=0.950 m2;d=2.50 cm=0.0250 m;T1=0∘C;T2=35.0∘C;t=1 day=24hours=86,400 s.
Identify the unknowns. We need to solve for the mass of the ice, m. We will also need to solve for the net heat transferred to melt the ice, Q. Determine which equations to use. The rate of heat transfer by conduction is given by
Q
t
=
k
A
(
T
2
−
T
1
)
d
Qt=kA(T2−T1)d
The heat is used to melt the ice: Q mLf.
Insert the known values:
Q
t
=
(
0.010
J/s
⋅
m
⋅
∘
C
)
(
0.950
m
2
)
(
35.0
∘
C
−
0
∘
C
)
0.0250
m
=
13.3
J/s
Qt=(0.010 J/s⋅m⋅∘C)(0.950 m2)(35.0∘C−0∘C)0.0250 m=13.3 J/s
Multiply the rate of heat transfer by the time (1 day = 86,400 s): Q =
(
Q
t
)
t
(Qt)t = (13.3 J/s)(86,400 s) = 1.15 × 106 J.
Set this equal to the heat transferred to melt the ice: Q = mLf. Solve for the mass m:
m
=
Q
L
f
=
1.15
×
10
6
J
334
×
10
3
J/kg
=
3.44
kg
m=QLf=1.15×106 J334×103 J/kg=3.44 kg
Discussion
The result of 3.44 kg, or about 7.6 lbs, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Inspecting the conductivities in Table 1 shows that Styrofoam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like Styrofoam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity.
Table 1. Thermal Conductivities of Common Substances[1]
Substance Thermal conductivity k (J/s⋅m⋅ºC)
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel iron 80
Steel (stainless) 14
Ice 2.2
Glass (average) 0.84
Concrete brick 0.84
Water 0.6
Fatty tissue (without blood) 0.2
Asbestos 0.16
Plasterboard 0.16
Wood 0.08–0.16
Snow (dry) 0.10
Cork 0.042
Glass wool 0.042
Wool 0.04
Down feathers 0.025
Air 0.023
Styrofoam 0.010