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bogdanovich [222]
3 years ago
9

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 th

e entropy of the two reservoirs changes. (Round the final answer to six decimal places.)
Chemistry
1 answer:
blagie [28]3 years ago
8 0

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.

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Answer:

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Explanation:

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