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2 years ago

A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.

1 answer:

6 0

The first law of thermodynamics says that the variation of internal energy of a system is given by:
$\Delta U = Q + W$
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
$\Delta U = -1275 J+855 J=-420 J$

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2 years ago

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2 years ago

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