An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C. 225 ∘C. During this process, 346 J of work was done by the system and its internal energy increased by 6085 J. 6085 J. What is the specific heat of the gas?

402 J / (kg ⁰C)

Explanation:

m = mass of the sample = 80.0 g = 0.080 kg

T₀ = Initial temperature of the sample = 25 ⁰C

T = Final temperature of the sample = 225 ⁰C

W = Amount of work done by the system = 346 J

U = Increase in internal energy = 6085 J

Q = Amount of heat given to the gas

Amount of heat is given as

Q = U + W

Q = 6085 + 346

Q = 6431 J

Heat received by sample is given as

Q = m c (T - T₀)

inserting the values

6431 = (0.080) c (225 - 25)

c = 402 J / (kg ⁰C)