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 6945 J. What is the specific heat of the gas?

Answer 1

we have relation dU = Q + W  

where dU = change in internal energy = 6945 J ,   W = work done = -346 J   ( since work done by system we take -ve sign) , we find Q

6945 = Q -346

Q = 7291 J = heat absorbed by gas

we have formula Q = m x S x dT    where m = mass of gas = 80g,   S = specific heat of gas ,

dT = change in temp == 225-25 = 200 c

substituting we egt 7291 J = 80g x ( S) x ( 200c)

S ( specific heat) = 0.4557 J/gC