An 80.0-gram sample of gas was heated from 25*C to 225*C. During the process, 346 J of work was done by the system and its internal energy increased by 8875. What is the specific heat of the gas?

Answer 1

So, I think to answer this question we will first want to solve for the heat (q) first, and then solve for specific heat of the gas.

(1) ∆U = q + w

∆U = change in internal energy = +8875 J (joules are assumed since you didn't provide units)

q = heat = ?

w = work = -346 J (negative since work was done BY the system ON the surroundings)

Solving for q, we have ...

q = ∆U - w

q = 8875 J - (-346 J)

q = 9221 H

We will now use the following equation to find the specific heat of the gas:

q = mC∆T

q = heat = 9221 J

m = mass of gas = 80.0 g

C = specific heat of gas = ?

∆T = change in temperature of the gas = 225º 25º = 200º

Solving for C, we have...

C = q / (m)(∆T) = 9221 J / (80.0 g)(200º)

C = 0.576 J / gº