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Heat Conduction

  • Thread starter pkossak
  • Start date
  • #1
52
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A styrofoam cooler (K = .030 W/mo C) has an average surface area of 0.465 m2 and an average thickness of 2.0 cm. About how long, in seconds will it take for 4.10 kg of ice at 0oC to melt in the cooler if the outside temperature is 28.0oC?

delta Q/delta t = (k*A*deltaT)/d
delta Q = 333 J/g (Lf) *4100 g = 1365300 J
(4186 J/kg*deltaT)*4.1 kg*28 C = 480552.8 J
1365300 J + 480552.8 J = 1845852.8 J = delta Q

1845852.8/delta t = ((.030 W/m C)*(0.465 m^2)*(28 C))/.02 m

Where did I go wrong?! :confused: Thanks a lot
 

Answers and Replies

  • #2
Fermat
Homework Helper
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I'm a bit vague on this stuff at the moment, but I would say that you went wrong when you raised the temp of water at 0ºC to water at 28ºC. (your third line)
This doesn't happen!

The temp difference (28 - 0) is just to determine the rate at which heat is being transferred from the outside of the cooler wall to the inside.
That heat is then used to melt 4.1 kg of ice at 0ºC into water at 0ºC.

Just skip the heat you added to raise the temp of water from 0 to 28.
 
  • #3
52
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I don't understand. That's still energy being used, shouldn't it be included?
 
  • #4
Fermat
Homework Helper
872
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But no water has its temp raised! That's why it shouldn't be included

Heat is transfered from the outside of the wall to the inside.
All of that heat is used to melt ice. That's all. No water is heated.

What you have is an ice-water mixture.
Ice can exist at 0ºC or below.
Water can exist at 0ºC and above.
If you have a water-ice mixture, then all of it must be at 0ºC since this is the only temp at which both ice and water can co-exist.
Since the temp of the mixture is always at 0ºC, then water can't have its temp raised. So no energy is involved in doing that.
 
  • #5
107
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You need to use the heat of fusion for water to determine how much energy is required to melt the ice. Then you find how long it takes the outside heat to permeate the container and melt the ice.
 

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