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If 425 joules are absorbed from the higher temperature reservoir and 295 joules of that is passed on to the lower temperature reservoir, how much work has been done by the system?

2 Answers

6 votes

Answer:

130 j

Step-by-step explanation:

User JoeKir
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3 votes

Answer: 130 J

Step-by-step explanation:

For the law of conservation of energy, the total energy absorbed by the hot reservoir must be equal to the sum of the energy passed to the cold reservoir and the work done by the system:


Q_H = Q_C + W

In this case, we have


Q_H = 425 J\\Q_C = 295 J

So, the work done by the system is


W=Q_H-Q_C=425 J-295 J=130 J

User Aditya Rewari
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