80.0 g 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 6905 J. What is the specific heat of the gas?

Answer 1

q = mC∆T

∆E = q + w where ∆E is change in internal energy; q = heat; w = work. Solving this for q, we have...

q = ∆E - w

∆E = +6905 J

w = -346 J (negative value because work was done BY the system, and not on it)

q = 6905 - (-346) = 7251 J. We can now use this in q = mC∆T

q = mC∆T

q = 7251 J

m = mass = 80.0 g

C = specific heat = ?

∆T = change in temperature = 225º - 25º = 200º

C = q / (m)∆T) = 7251 J / (80.0 g)(200º)

C = 0.453 J/gº