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 6485 j. what is the specific heat of the gas? c = j/(g·°c)

Explanation:

Step1: Use the first law of thermodynamics

The first law of thermodynamics is $\Delta U = q+w$. We know that $\Delta U=6485\ J$ and $w = - 346\ J$ (work done by the system). So, we can find $q$:
$q=\Delta U - w$
$q=6485\ J-(-346\ J)=6485 + 346=6831\ J$

Step2: Use the heat formula

The heat formula is $q = mc\Delta T$. Here, $m = 80.0\ g$, $\Delta T=T_{2}-T_{1}=225 - 25=200^{\circ}C$ and $q = 6831\ J$. We can solve for $c$:
$c=\frac{q}{m\Delta T}$
$c=\frac{6831\ J}{80.0\ g\times200^{\circ}C}$
$c=\frac{6831}{16000}\ J/(g\cdot^{\circ}C)\approx0.427\ J/(g\cdot^{\circ}C)$

Answer:

$0.427\ J/(g\cdot^{\circ}C)$