Question

Heat is transferred at a rate of 2 kW from a hot reservoir at 775 K to a cold reservoir at 300 K. Calculate the rate at which the entropy of the two reservoirs changes. (Round the final answer to six decimal places.)

Answer 1

Explanation:

The given data is as follows.

           Q = 2 kW,        $T_{1}$ = 775 K

       $T_{2}$ = 300 K

The relation between entropy and heat energy is as follows.

             $\Delta S = \frac{Q}{\Delta T}$

Therefore, calculate the entropy at each temperature as follows.

                 $S_{1} = \frac{Q}{T_{1}}$

                            = $\frac{2 kW}{775 K}$

                            = $2.5 \times 10^{-3}$ kW/K

Also,        $S_{2} = \frac{Q}{T_{2}}$

                            = $\frac{2 kW}{300 K}$

                            = $6.6 \times 10^{-3}$ kW/K

Hence, the change in entropy will be calculated as follows.

          $\Delta S = S_{2} - S_{1}$

                     = $(6.6 \times 10^{-3} - 2.5 \times 10^{-3})$ kW/K

                     = $4.1 \times 10^{-3}$ kW/K

or,                 = 0.0041 kW/K

Thus, we can conclude that the rate at which the entropy of the two reservoirs changes is 0.0041 kW/K.