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?

Question

aaronq · Answer

internal energy, U, is equal to the work done or by the system, plus the heat of the system:

   $\Delta U=q+w$

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 $C_p$.
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remember that $ΔT=T_f−T_i$

so the equation, really, is: $q=m∗Cp∗(T_f−T_i)$
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$ 8185 J= -346 J + [m∗Cp∗(T_f−T_i)]$

plug in the rest of your values and solve for $C_p$