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?

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.


remember that ΔT=Tf−Ti

so the equation, really, is: q=m∗Cp∗(Tf−Ti)


185J=−346J+[m∗Cp∗(Tf−Ti)]

plug in the rest of your values and solve for Cp

Depotion can i think