Question

A cube of side $5 \,cm$ made of iron and having a mass of $1500\, g$ is heated from $25^{\circ} C$ to $400^{\circ} C$. The specific heat for iron is $0.12 \,cal / g ^{\circ} C$ and the coefficient of volume expansion is $3.5 \times 10^{-5} /{ }^{\circ} C$, the change in the internal energy of the cube is $\left( atm\right.$ pressure $=1 \times 10^5 \,N / m ^2$ )

• A. $320 \,kJ$
• B. $282 \,kJ$
• C. $141\, kJ$
• D. $4023\, kJ$

Answer 1

Answer: (B)

$Q =m C \Delta T=1.5 \times 0.12 \times 4200 \times(400-25) $
$=2.83 \times 10^5 \, J $
$W =P(\Delta V)=P(V \gamma \Delta T) $
$=10^5 \times\left(5 \times 10^{-2}\right)^3 \times 3.5 \times 10^{-5} \times 375=0.164\, J$
Thus $Q=\Delta U+W$
or $2.83 \times 10^5=\Delta U+0.164 ; \Delta U =282\, kJ$