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s344n2d4d5 [400]
3 years ago
8

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 7585 J. What is the specific heat of the gas?
Chemistry
1 answer:
german3 years ago
5 0
Hello!

First you need to calculate q 
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>

<span>q = m x c x delta T </span>

<span>7211 J = 80.0 g x c x (225-25) °C </span>

<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
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