Question

5- The inside of a Carnot refrigerator is maintained at a temperature of 277 K, while the temperature in the kitchen is 303 K. Using 2159 J of work, how much heat can this refrigerator remove from its inside compartment

Answer 1

Answer:

The heat is  $H = 23001.65 \ J$    

Explanation:

From the question we are told that

    The inside temperature is  $T_i = 277 \ K$

     Te kitchen temperature is  $T_k = 303 \ K$

      The  workdone is  $W = 2159 \ J$

     

The  coefficient of performance  of the refrigerator is mathematically represented as

           $COP = \frac{1}{\frac{T_k}{T_i} -1}$

substituting values  

        $COP = \frac{1}{\frac{303}{277} -1}$

       $COP = \frac{277}{26}$

        $COP = 10.65$

Now  this   coefficient of performance  of the refrigerator  can also be represented mathematically as

         $COP = \frac{H}{W}$

Where  H is the heat which the refrigerator removes from the inside component

     So  

            $H = COP * W$

substituting values  

            $H = 10.65 * 2159$            

            $H = 23001.65 \ J$