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klio [65]
3 years ago
5

The internal energy of a system changes because the system gains 160 J of heat and performs 309 J of work. In returning to its i

nitial state, the system loses 106 J of heat. During this return process, what work is involved, and is the work done by the system or on the system?
Physics
1 answer:
o-na [289]3 years ago
5 0

Answer:

w=255

Explanation:

The change in internal energy is given by the first law:

ΔE = Q - w

where ΔE is the change in internal energy of the system

q is the heat added to the system

w is the work done *by* the system on the surroundings

So, for the first phase of this process:

ΔE = Q - w

Q=160J

w=309J

ΔE = 160J - 309J = -149J

To bring the system back to its initial state after this, the internal energy must change by +149J (the system myst gain back the 149 J of energy it lost).  We are told that the system loses 106 J of heat in returning to its initial state, so the work involved is given by:

ΔE = Q - w

+149J = -106J - w

255J = -w

w = -255J

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