Thermodynamics- Refrigerator Problem

[jessedevin]

Homework Statement

A refrigerator is operated by a 0.25 hp (1 hp=746 watts) motor. If the interior is to be maintained at 2.00 degrees Celsius and the room temperature of the room is 35 degrees C, what is the maximum heat leak in watts that can be tolerated? Assume that the coefficient of preformance is 50% of the maximum theoretical vale. What happens if the leak is greater than your calculated maximum value.

Homework Equations

[tex]\eta[/tex]= T_cold/(T_hot-T_cold)
[tex]\eta[/tex]=Q_cold/w
Q_hot=Q_cold+W.

The Attempt at a Solution

What I am first doing is finding the coefficient of performance [tex]\eta[/tex]
[tex]\eta[/tex]=375K/(408K-375K)=11.36
Then it says that the coefficent of preformance is 50% of the the max theoretical value, so [tex]\eta[/tex]= 11.36/2=5.68
Then I find the rate of Q_cold = w*[tex]\eta[/tex]
Q_cold=(746w/4)(5.68)=1059.66w
Lastly I used Q_hot=Q_cold+W,
Q_hot=1059.66w+ 746w/4=1246.16w

But the answer in my book says its 779 watts, so did I miss a step or do something wrong. And what happens if the leak is greater than your calculated maximum value?

 

[Andrew Mason]

First, check your conversion to Kelvin temperature. It is not correct.

Second, your method is correct except that you do not calculate Qh. You use Qc, since this is the heat that has to be removed.

What do you think happens if the refrigerator is not able to remove the heat as fast as it is entering the refrigerator?

AM

 

[tomcue]

Your approach to solving the problem seems correct. However, there may be a mistake in your calculation of Qcold. It should be Qcold = w/eta = (746 W)/(4*5.68) = 32.99 W. This means that the maximum heat leak that can be tolerated is 32.99 W.

If the heat leak is greater than this calculated value, the refrigerator will not be able to maintain the desired temperature of 2.00 degrees Celsius. The temperature inside the refrigerator will start to increase and eventually reach the room temperature of 35 degrees Celsius. This will result in spoilage of food and reduced efficiency of the refrigerator. In order to prevent this, the heat leak needs to be reduced or the efficiency of the refrigerator needs to be improved.