Question

An 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 9185 J. What is the specific heat of the gas?

Answer 1

Answer:

$C=0.596\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to the first law of thermodynamics it is possible to compute the gained heat by the water as shown below:

$Q=\Delta U+W\\\\Q=9185J+346J=9531 J$

Thus, it is possible to solve for the specific heat of the gas as shown below:

$Q=mC\Delta T\\\\C=\frac{Q}{m\Delta T}=\frac{9531J}{80.0g*(225-25)\°C}\\\\C=0.596\frac{J}{g\°C}$

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