Asked Dec 3, 2019

Suppose a system returns to its original internal energy after the following changes. In step 1, 25 J of work is done on the system as it releases 37 J of heat energy. If, in step 2, the system performs 12 J of work, what is the value of q?

a. 0 J

b. -24 J

c. +24 J

d. -49 J

e. +49 J


Expert Answer

Step 1

It is given that in first step system releases 37 J of energy which means q = -37 and the work is done on the system is 25 J.

Step 2

Hence, the change in energy of surrounding for first step is,


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AE, q + W1 -((-37)+(25)J =-12 J

Step 3

In second step, it is given that system performs 12 J of work, which means the sign of work done will ...


Image Transcriptionclose

AE, 2W2 AE2 (42-12)J


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Chemical Thermodynamics

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