Question

A heat engine operates between a hih temp reservoir at 610K and low temp reservoir at 320K. In one cycle, the engine absorbs 6400J from the high temp reservoir and does 2200 J of work. What is the net change in entropy as a result of this cycle?

Answer 1

To solve this problem it is necessary to apply the concepts related to entropy. Entropy can be defined as the change between heat energy and body temperature. For this case we will analyze the entropy in the cold body and later in the hot body. From there we will find the entropy difference

Entropy can be defined as,

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

Here,

Q = Heat Flow

T = Temperature

The entropy of the hot reservoir goes down by

$S_1 = \frac{6400J}{610K}$

The entropy of the "cool" reservoir increases by

$S_2 = \frac{4200J}{320K}$

$\Delta S = S_2 -S_1$

$\Delta S = \frac{4200J}{320K}-\frac{6400J}{610K}$

$\Delta S = 13.125-10.492$

$\Delta S = 2.633J/K$

As expected, entropy increased in this process; the entropy lost by the hot side was less than the entropy gained by the cool side.

Answer 2

Answer:

the net change in entropy as a result of this cycle is 2.63 J/K

Explanation:

given information:

high temperature, $T_{H}$ = 610 K

low temperature, $T_{C}$ = 320 K

heat absorbed, $Q_{H}$ = 6400 J

work, W = 2200 J

to calculate the net change in entropy as a result of this cycle, we can use the following formula:

ΔS = $\frac{Q_{H} }{T_{H} } +\frac{Q_{C} }{T_{C} }$ and W = $Q_{H}$ - $Q_{C}$

where

ΔS = the change of entropy

$T_{H}$ = the high temperature

$T_{C}$ = low temperature

$Q_{H}$ = absorbed heat

$Q_{C}$ = released heat

first, we determine the $Q_{C}$

W = $Q_{H}$ - $Q_{C}$

$Q_{C}$ = 6400 - 2200

     = 4200 J

therefore,

ΔS = $\frac{Q_{H} }{T_{H} } +\frac{Q_{C} }{T_{C} }$

     = $-\frac{6400}{610} + \frac{4200}{320}$

     = - 10.49 + 13.125

     = 2.63 J/K