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

Step-by-step 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 = (Q)/(\Delta T)`

Therefore, calculate the entropy at each temperature as follows.

`S_(1) = (Q)/(T_(1))`

=
`(2 kW)/(775 K)`

=
`2.5 * 10^(-3)` kW/K

Also,
`S_(2) = (Q)/(T_(2))`

=
`(2 kW)/(300 K)`

=
`6.6 * 10^(-3)` kW/K

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

`\Delta S = S_(2) - S_(1)`

=
`(6.6 * 10^(-3) - 2.5 * 10^(-3))` kW/K

=
`4.1 * 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.