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 7905 J. What is the specific heat of the gas?

Question

aaronq · Answer

internal energy is given by: U = q + w

the heat can be found with: $q=m*C_p*\Delta T$

substituting (2) into (1):

$U = (m*C_p*\Delta T) + w$

aaronq · Answer

solve for specific heat capacity $(C_p)$.

anonymous · Answer

I'm not sure of what to put where..I know w is -346 and m 80

aaronq · Answer

"internal energy increased by 7905 J"

$\Delta U=7905 J=m*C_p*(T_f-T_i)+w$

because $\Delta T=T_f-T_i$

anonymous · Answer

I get .505 but I think that is wrong

anonymous · Answer

so I got 7905J=80g*Cp*200+-346 but .505 is wrong

aaronq · Answer

i got Cp=0.5156875

anonymous · Answer

that is the correct answer but how did you get that...also thank you

aaronq · Answer

No problem, i used that same equation, you probably messed up in the arithmetic because you were pretty close.