   Chapter 15, Problem 4PE

Chapter
Section
Textbook Problem

# What is the change in internal energy of a system which does 4.50 × 10 5 J of work while 3.00 × 10 6 J of heat transfer occurs into the system, and 8.00 × 10 6 J of heat transfer occurs to the environment?

To determine

The change in internal energy of the system.

Explanation

Given info:

Heat transfer into the system, Qin=3.00×106 J.

Heat transfer from the system to the environment, Qout=8.00×106 J.

Net heat transfer =ΔQ=?

Work is done by the system W=4.50×105 J (positive since work is being done by the system)

Change in internal energy =ΔU=?

Formula Used:

Net Heat transfer is given as

ΔQ=Qin+Qout

According to the first law of thermodynamics, we have

ΔQ=ΔU+W

Calculation:

Net heat transfer is given as,

ΔQ=Qin+Qout

Inserting the values

ΔQ=3

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