How much does a light heat the air around it?
During a period of $\Delta t = 1\ \mathrm h =3600\ \mathrm s$, the light bulb converts electric power of $P = 60\ \mathrm W$ to light and heat. Assuming that none of this energy escapes from the room, the entire power is eventually converted to heat:

$$\begin{align}
Q &amp;= P \cdot \Delta t \\
&amp;= 60\ \mathrm W \times 3600\ \mathrm s \\
&amp;= 216\ \mathrm{kJ}
\end{align}$$

According to REFPROP (NIST Standard Reference Database 23, Version 9.0), the relevant parameter values of air at the assumed initial temperature of $T_0 = 20\ \mathrm{^\circ C}$ and pressure $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$ are a density of $\rho_0 = 1.2046\ \mathrm{kg/m^3}$ and a specific heat capacity at constant pressure of $c_{p,0} = 1.0061\ \mathrm{kJ/(kg\ K)}$.

The initial volume is given as $V_0 = 85\ \mathrm{m^3}$. The corresponding mass is

$$\begin{align}
m &amp;= V_0 \cdot \rho_0 \\
&amp;= 85\ \mathrm{m^3} \times 1.2046\ \mathrm{kg/m^3} \\
&amp;= 102.391\ \mathrm{kg}
\end{align}$$

Ignoring the heat capacity of the walls and inventory, the temperature of the air is approximately increased by

$$\begin{align}
\Delta T &amp;= \frac QC \\
&amp;= \frac Q{m \cdot c_{p,0}} \\
&amp;= \frac{216\ \mathrm{kJ}}{102.391\ \mathrm{kg} \times 1.0061\ \mathrm{kJ/(kg\ K)}} \\
&amp;= 2.1\ \mathrm{K}
\end{align}$$

By way of comparison, a better result may be obtained using the enthalpy of air, since the heat capacity of air is not constant. In this case, however, the difference is insignificant because the temperature change is small. The initial specific enthalpy of air is $h_0 = 419.40\ \mathrm{kJ/kg}$. Thus, the enthalpy is

$$\begin{align}
H_0 &amp;= m \cdot h_0 \\
&amp;= 102.391\ \mathrm{kg} \times 419.40\ \mathrm{kJ/kg} \\
&amp;= 42\,943\ \mathrm{kJ}
\end{align}$$

At constant pressure, the change in enthalpy corresponds to the heat transferred to the air:

$$\begin{align}
H_1 &amp;= H_0 + \Delta H \\
&amp;= H_0 + Q \\
&amp;= 42\,943\ \mathrm{kJ} + 216\ \mathrm{kJ} \\
&amp;= 43\,159\ \mathrm{kJ}
\end{align}$$

Thus, the final specific enthalpy is

$$\begin{align}
h_1 &amp;= \frac{H_1}{m} \\
&amp;= \frac{43\,159\ \mathrm{kJ}}{102.391\ \mathrm{kg}} \\
&amp;= 421.51\ \mathrm{kJ/kg}
\end{align}$$

At a pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, this specific enthalpy corresponds to a temperature of $T = 22.1\ \mathrm{^\circ C}$.

Note that, at the final temperature of $T = 22.1\ \mathrm{^\circ C}$ and the pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, the new density of air amounts to $\rho_1 = 1.1960\ \mathrm{kg/m^3}$. Thus, the new volume is

$$\begin{align}
V_1 &amp;= \frac{m}{\rho_1} \\
&amp;= \frac{102.391\ \mathrm{kg}}{1.1960\ \mathrm{kg/m^3}} \\
&amp;= 85.6\ \mathrm{m^3}
\end{align}$$

Therefore, a small amount of air must escape from the room.

Finally, if the heat from a $60\ \mathrm W$ light bulb really is a problem, you might want to leave the room since the human body generates about $100\ \mathrm W$.