91Ó°ÊÓ

When an object of area A at temperature\({{\bf{T}}_{\bf{1}}}\)is placed in the surroundings at temperature\({{\bf{T}}_{\bf{2}}}\), it not only radiates energy but also absorbs the energy radiated by other objects.

The net rate of radiant heat flow from the object is given as:

\(\frac{Q}{t} = \varepsilon \sigma A\left( {T_1^4 - T_2^4} \right)\) … (i)

Here, \(\varepsilon \) is the emissivity, A is the area, and \(\sigma \) is known as the Stefan-Boltzmann constant whose value is \(\sigma = 5.67 \times {10^{ - 8}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^4}\).

The power of the light bulb on the glass surface is\(P = 75\;{\rm{W}}\).

The temperature is\(T = 75^\circ {\rm{C}}\).

The room temperature is\({T_{\rm{r}}} = 18^\circ {\rm{C}}\).

The power of the other light bulb is\({P_0} = 150\;{\rm{W}}\).

The percentage of heat emitted is\(\eta = 90\% \).

The relation of heat transfer is given by:

\(\frac{Q}{t} = \varepsilon \sigma A\left( {{T^4} - {T_{\rm{r}}}^4} \right)\)

Substitute the values in the above expression.

\(\begin{aligned}{c}\left( {0.9} \right)\left( {75\;{\rm{W}}} \right) = \varepsilon \sigma A\left( {{{\left( {75^\circ {\rm{C}} + 273\;{\rm{K}}} \right)}^4} - {{\left( {15^\circ {\rm{C}} + 273\;{\rm{K}}} \right)}^4}} \right)\\\varepsilon \sigma A = 9.63 \times {10^{ - 9}}\,{\rm{W/}}{{\rm{K}}^{\rm{4}}}\end{aligned}\)

The relation to find the temperature is given by:

\({\left( {\frac{Q}{t}} \right)^\prime } = \varepsilon \sigma A\left( {{{T'}^4} - {T_{\rm{r}}}^4} \right)\)

Here,\(T'\)is the required temperature and\({\left( {\frac{Q}{t}} \right)^\prime }\)is the power of the other light bulb.

Substitute the values in the above expression.

\(\begin{aligned}{c}150\;{\rm{W}} = \left( {9.63 \times {{10}^{ - 9}}\,{\rm{W/}}{{\rm{K}}^{\rm{4}}}} \right)\left( {{{T'}^4} - {{\left( {15^\circ {\rm{C}} + 273\;{\rm{K}}} \right)}^4}} \right)\\T' \approx 387\;{\rm{K}}\\ = \left( {387 - 273} \right)^\circ {\rm{C}}\\ = 114^\circ {\rm{C}}\end{aligned}\)

Thus, the temperature of a light bulb is \(114^\circ {\rm{C}}\).