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 7585 J. What is the specific heat of the gas?

Answer 1

Hello!

First you need to calculate q
delta U is change in internal energy

delta U = q + w
q is heat and w work done
here work was done by the system means energy leaving the system so w is negative

delta U = q + w

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ

q = m x c x delta T

7211 J = 80.0 g x c x (225-25) °C

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)