First slide

First law of thermo dynamics

Question

During a process, work equivalent to 400J is done on a system, which gives out of 125J of heat energy. The change in internal energy is

Easy

Solution

W = +400J….work done on the system…….positive sign
q = -125J……heat liberated by the system…negative sign
ΔE = ?
ΔE = -125+400=275 J

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