Question

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 8185 J. What is the specific heat of the gas?

Answer 1

Internal energy, U, is equal to the work done or by the system, plus the heat of the system: ΔU=q+w
in the question they tell you the work done by the system, and the internal energy:
8185 J= -346 J + q work is negative because it was done BY the system.
substitute in: q=m∗Cp∗ΔT and solve for Cp.

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remember that ΔT=Tf−Ti

so the equation, really, is: q=m∗Cp∗(Tf−Ti)
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185J=−346J+[m∗Cp∗(Tf−Ti)]

plug in the rest of your values and solve for Cp