Question

A metal ball of surface area $200$ $c{m}^2$ and temperature ${527}^{\circ }C$ is surrounded by a vessel at ${27}^{\circ }C$. If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is approximately: $\sigma =5.67\times {10}^{-8}\frac{joule}{m^2\times sec\times K^2}$

• A. 108 joule
• B. 168 joule
• C. 182 joule
• D. 192 joule

Answer 1

The correct option is C 182 joule

From stefan's law we know that,rate of loss of heat is given as,
$\frac{Q}{t}=\sigma A({T_1}^4-{T_2}^4)\times e$
so,$T_1=273+527K=800K$
$T_2=273+27K=300K$
and $A=200\times {10}^{-4}{m}^2$
so,$\frac{Q}{t}=5.67\times {10}^{-8}\times 2\times {10}^{-2}\left[\right({800}^4)-({300}^4\left)\right]\times \left(0.4\right)=\left(182.12\right)Joule$