# Question 5992a

Jul 31, 2016

We can try to solve the problem using Stefan-Boltzman law of radiation.According to this law we have the follwing equation.

$\rho = A \sigma e \left({T}^{4} - {T}_{c}^{4}\right)$

$\text{Where}$

$\rho \to \text{Net radiated power}$
$A \to \text{Radiaditing area}$
$\sigma \to \text{Stefan's constant}$
$e \to \text{Emissivity}$
$T \to \text{Temperature of radiating surface}$
${T}_{c} \to \text{Temperature of surrounding }$

So $\rho$ will be proportional to the rate of eletrical energy spent or wattage $E$ of the lamp .
SinceA,sigma,e=90% (given) are remaining same in two cases.

So $\rho \propto E \implies \left({T}^{4} - {T}_{c}^{4}\right) \propto E$

In our problem

For first lamp

${E}_{1} \to \text{Electrical power} = 60 W$
${T}_{1} \to \text{Temperature of lamp} = {65}^{\circ} C = \left(65 + 273\right) K = 338 K$
${T}_{c} \to \text{Temperature of surrounding} = {18}^{\circ} C = \left(18 + 273\right) K = 291 K$

For 2nd lamp

${E}_{2} \to \text{Electrical power} = 150 W$
T_2->"Temperature of lamp"=?#
${T}_{c} \to \text{Temperature of surrounding} = {18}^{\circ} C = 291 K$

So we can write

$\frac{{T}_{2}^{4} - {T}_{c}^{4}}{{T}_{1}^{4} - {T}_{c}^{4}} = {E}_{2} / {E}_{1} = \frac{150}{60} = 2.5$

$\implies \frac{{T}_{2}^{4} - {291}^{4}}{{338}^{4} - {291}^{4}} = 2.5$

$\implies {T}_{2}^{4} = \left({338}^{4} - {291}^{4}\right) \times 2.5 + {291}^{4}$

${T}_{2} = 384.6 K = {\left(384.6 - 273\right)}^{\circ} C = {111.6}^{\circ} C$