Question

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extract energy by heat transfer at a rate of 3050 J/s from the freezer contents, which are at a temperature of 270 K. Determine if this claim is real considering an ambient temperature of 293 K. (a) Can the freezer operation in such conditions

Answer 1

Answer:

The inventors  claim is not real

a)  No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

     The  power input is  $P_i = 0.25 kW = 0.25 *10^{3} \ W$

      The  rate of heat transfer $J = 3050 J/s$

       The temperature of the freezer content is $T = 270 \ K$

       The  ambient temperature is  $T_a = 293 \ K$

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

      $COP = \frac{T }{Ta - T}$

substituting values

     $COP = \frac{270 }{293 - 270}$

     $COP =11.7$

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

       $COP = \frac{J}{P_i}$

substituting values

       $COP = \frac{3050}{0.25 *10^{3}}$

       $COP = 12.2$

Now given that the COP  of an ideal refrigerator is  less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP