The heat energy that must be supplied to 14 gram of nitrogen at room temperature to raise its temperature by $48^{\circ} \mathrm{C}$ at constant pressure is (Molecular weight of nitrogen $=28, R=$ gas constant, $C_p=\frac{7}{2} R$ for diatomic gas)

Heat energy supplied at constant pressure is governed by $Q_p = n C_p \Delta T$, where $n$ is the number of moles, $C_p$ is molar specific heat at constant pressure, and $\Delta T$ is the temperature change.

For diatomic nitrogen, molar specific heat at constant pressure is $C_p = \frac{7}{2} R$. Given a mass of $14$ g and molecular weight $28$ g/mol, the number of moles is $n = \frac{14}{28} = 0.5$ mol. With $\Delta T = 48^{\circ}\mathrm{C}$:

$Q_p = (0.5) \cdot \left(\frac{7}{2} R\right) \cdot 48 = 0.5 \cdot \frac{7}{2} \cdot 48 R = \frac{7}{4} \cdot 48 R = 7 \cdot 12 R = 84 R$.

The result is $84 R$, which corresponds to option B.