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Spread of energy in fridge?

  • Thread starter pivoxa15
  • Start date
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1. Homework Statement
A measure of 0.5kg of water is placed in a 0.2kg aluminium tray, both initially at 288k and placed in a fridge that is kept at 268k. How long will it take to freeze the water and lower the temperature to –5celcius if the rate of removal of heat energy by the fridge is 2kJ each minute?


2. Homework Equations
specific heat of water =4.2*10^3 J/kg/K
Specific latent heat of Fusion of water = 3.4*10^5 J/kg


3. The Attempt at a Solution

I got that the total energy removed from the 0.5kg of water needs to be 6366J so a total of 106.1minutes is needed and the answers at the back of the book gave this figure as well. My question is what about the aluminium tray? Energy removal from the fridge removes heat randomly so of the 106.1 minutes it doesn’t always remove heat from the water but should also remove heat from the aluminium as well? So it should take more than 106.1 minutes in reality wouldn’t it? How about double that since on average, in 50% of the time heat is removed from the water and the other 50% of the time, heat is removed from the aluminium. How would you go about it? I find it hard to believe that the book suggests the fridge will only remove heat from the water and none from the tray.
 

Astronuc

Staff Emeritus
Science Advisor
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Determine the heat content of the Al tray. Is it similar or negligble to that of the water. Also compare the thermal conductivity of Al with water. The greater the thermal conductivity, the easier (faster) the heating or cooling. If Al cools more rapidly then some of the heat of the water passes through the Al tray during the cooling process.
 
2,234
1
Determine the heat content of the Al tray. Is it similar or negligble to that of the water. Also compare the thermal conductivity of Al with water. The greater the thermal conductivity, the easier (faster) the heating or cooling. If Al cools more rapidly then some of the heat of the water passes through the Al tray during the cooling process.
How do I determine the heat content of the Al tray if I am only given its temperture?

The secific heat of Al is roughly an order of magnitude smaller so the aluminium will cool much faster than the water. If we assume the Al to be -5celcius as well than it takes an order of magnitude less energy to cool it to that temperture so during the cooling process the Al will always be cooler than the water so most of the cooling by the fridge is done to the water. So if it takes 106 minutes to cool the water to that temperture and only a minute or so is needed to cool Al to that temperture so a total of about 107minutes is needed. The answers should have taken this into account even though small.
 

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