Question

Consider a block of copper of radius 5 cm. Its outer surface is coated black. How much time is required for the block to cool down from 1000K to 300K. Density of copper = $9\times {10}^{3}kg/m^3$ and specific heat = 4 kJ/kg.

• A. $127\times {10}^{3}sec$
• B. $27\times {10}^{3}sec$
• C. $12\times {10}^{3}sec$
• D. $1.0\times {10}^{3}sec$

Answer 1

The correct option is A $127\times {10}^{3}sec$

When the copper block is coated black, it behaves a perfectly black body. The rate of heat energy radiated out at any instant is given by
$\frac{dQ}{dt}=\sigma A{T}^4\dots \dots \left(1\right)$
where A is the surface area and T is the temperature of black body at that instant. Here it should be remembered that T changes continuously with time as black body cools.
The heat given out by a block of mass m and specific heat c, for unit of temperature is given by,
$\frac{dQ}{dT}=mc\dots \dots \left(2\right)\therefore \frac{dT}{dt}=-\frac{\sigma A}{mc}{T}^4\dots \dots \left(3\right)$
Negative sign is used to show that temperature falls as time increases.
From eq. (3),
$dt=-\frac{mc}{\sigma A}\frac{dT}{T^4}\dots \dots \left(4\right)$
Hence, time required for the block to cool down from $T_1$ to $T_2$ is given by
${}_{T_0}^{T}dt=-{\frac{mc}{\sigma A}}_{T_0}^T\frac{dT}{T^4}$
Integrating, we get
$t=\frac{mc}{3\sigma A}\frac{1}{T_2^3}-\frac{1}{T_1^3}=\frac{mc}{3\sigma A}\frac{T_1^3-T_2^3}{T_1^3{T}_2^3}$
If $\rho$ be the density of copper and r be the radius of sphere, then
$m=\frac{4}{3}\pi r^3\rho andA=4\pi r^2,\therefore t=\frac{r\pi c}{9\sigma }\frac{T_1^3-T_2^3}{T_1^3{T}_2^3}Herer=5cm=5\times {10}^{-2}m,r=9\times {10}^{3}kg/m^3,c=4\times {10}^{3}J/kg,T_1=1000={10}^{3}K,T_2=300Kand\sigma =5.67\times {10}^{-8}J/m^{2}s{K}^4$

Substituting the given data in the obtained above expression for time gives the answer $127\times {10}^{3}sec$.