Temperature related doubts

Question

if a refrigerator works between the temperature 4c &30c and 600 calories of heat is released then how to calculate work in joules ?

Answer 1

Hi Prashant!

 

I assume that this refrigerator is operating at its Carnot performance, so that the temperatures between which it is operating can be used to calculate the refrigerator's coefficient of performance (COP).  Otherwise, I am not sure how to answer the question.

 

Refrigerators pull pull heat from a lower temperature region (inside) and shove it into a higher temperature region (outside).  This is opposite the way heat naturally flows, so it has to do work to push the heat into the higher temperature outside region.  The relationship for this is:

 

Q_C + W = Q_H

 

where

 

Q_C = amount of heat taken from lower temperature region

W = work done to shove heat into higher temperature region

Q_H = amount of heat expelled into higher temperature region

 

This can be arranged into:

 

Q_C = Q_H - W

 

This equation means the refrigerator removes some heat from inside, adds work to it, and then expels it outside.  It has to do work to shove the heat in a way it does not naturally want to go.

 

[NOTE: in a refrigerator, this work is usually done by the "compressor" unit, whose motor can often be heard whirring on the back of refrigerators as they are maintaining their internal temperature]

 

The coefficient of performance (COP) of a refrigerator, a measure of how well it keeps the inside cool, is determined by the ratio:

 

COP = Q_C/W

 

This means that, if one refrigerator can remove more heat from the inside while doing the same amount of work as another refrigerator, it is more efficient.  The more heat the refrigerator can remove from its inside using a given amount of work, the more efficient it is.

 

[NOTE: In practice, this efficiency is often called the "coefficient of performance" or COP]

 

The maximum refrigerator COP, in terms of the temperatures between which it operates (T_C and T_H), is

 

COP_max = T_C/(T_H-T_C)

 

where

 

T_C = temperature inside the refrigerator

T_H = temperature outside the refrigerator

 

This is the Carnot COP for a refrigerator.  The temperatures in this formula must be on an absolute scale (often Kelvin).

 

So, the refrigerator in this problem operates between 4 C and 30 C. In kelvins, this is:

 

T_C = 4 + 273.15 = 277.15 K

T_H = 30 + 273.15 = 303.15 K

 

So:

 

COP_max = 277.15 K/(303.15 K - 277.15 K) =  10.66

 

Thus:

 

10.66 = Q_C/W = (Q_H - W)/W = Q_H/W - 1

 

So:

 

Q_H/W = 10.66 + 1 = 11.66

 

W = Q_H/11.66

 

Q_H = 600 cal * 4.184 J/cal = 2510.4 J

 

Thus:

 

W = 2510.4 J/11.66 = 215.3 J

 

I hope this helps!  But, if you have any more questions about this or want to look at any steps more closely, just let me know.