Question

Find the change in internal energy in joule when $10g$ of air heated from ${30}^{\circ }C$ to ${40}^{\circ }C$
$(c_v=0.172kcal/kg/KJ=4200J/kcal)$

• A. $62.24J$
• B. $72.24J$
• C. $52.24J$
• D. $82.24J$

Answer 1

The correct option is B $72.24J$

Given,

$T_1={30}^{0}C=30+273=303K$

$T_2={40}^{0}C=40+273=313K$

$m=10g={10}^{-2}kg$

$\Delta V=0m^3$

$C_v=0.172Kcal$

Change in Heat energy,

$\Delta Q=m{C}_v(T_2-T_1)$

$\Delta Q={10}^{-2}\times 0.172\times 4200\times (313-303)$

$\Delta Q=0.172\times 42\times 10$

$\Delta Q=72.24J$

Work done, $W=P.\Delta V=0J$

From first law of thermodynamics,

$\Delta Q=\Delta U+W$

$\Delta U=\Delta Q$

$\Delta U=72.24J$

The correct option is B.