Originally Posted by **studiot** If the gas is heated at constant pressure, it must expand.
So it must do work against the surroundings.
So the heat added must equal the gain in internal energy of the gas to the new temperature plus the work done.
IOW not all the heat added goes to increasing the temperature. |

Yes, that's correct. That's why you must use $\displaystyle c_p$ in the formula for the heat transfer in the problem outlined above to get the correct answer. If you use $\displaystyle c_p$ in the heat transfer formula, you get the correct answer because the work done by the gas is factored out of the problem.

Note that if you know $\displaystyle c_v$, you can work out the change in temperature that would have occurred if the heat transfer was made instead at constant volume and therefore with no energy going into work done:

$\displaystyle T_2 = T_1 + \frac{Q}{mc_v} = 21 + \frac{4400}{0.76 \times 312} = 39.56$ degrees Celsius.

It's higher