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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 (which was stationary), and its internal energy increased by 7765 J. What is the specific heat capacity of the gas?

1 Answer

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Step 1: calculate q (charge)
delta U is change in internal energy

Using the formula:
delta U = q + w

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


q = mc x delta T
Step 2) Calculate C.

7211 J = 80.0 g x c x (225-25) °C

c = 0.451 J /g °C
User Florent Ferry
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