Chemistry: 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.)

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Heat is transferred at a rate of 2 kw from a hot reservoir

aladuke79

10.01.2022

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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