Question

A metal ball of emissivity 4/7 and surface area 100 $cm^2$ is at a temperature of $127^{\circ}C$. If the temperature of the surroundings is $27^{\circ}C$, then the rate of loss of heat of the ball is:

• A. 2.835 W
• B. 22.68 W
• C. 5.67 W
• D. 11.34 W

Hint:

Always convert temperatures to Kelvin ($K = ^{\circ}C + 273$) before using Stefan-Boltzmann!

Answer 1

The Correct Option is B

Step 1: The law for heat radiation.
A hot body loses heat by radiation following the Stefan-Boltzmann law. The net rate of heat loss to cooler surroundings is \[ P = e\,\sigma\,A\,\left(T^4 - T_0^4\right) \] where $e$ is emissivity, $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$, $A$ is the area, $T$ the body temperature and $T_0$ the surroundings.

Step 2: Convert the data.
Emissivity $e = \dfrac{4}{7}$. Area $A = 100\ \text{cm}^2 = 100\times10^{-4} = 10^{-2}\ \text{m}^2$. Temperatures in kelvin: $T = 127 + 273 = 400\ \text{K}$ and $T_0 = 27 + 273 = 300\ \text{K}$.

Step 3: Work out the temperature term.
\[ T^4 - T_0^4 = 400^4 - 300^4 = 256\times10^8 - 81\times10^8 = 175\times10^8\ \text{K}^4 \]

Step 4: Plug everything in.
\[ P = \frac{4}{7}\times(5.67\times10^{-8})\times10^{-2}\times175\times10^8 \] Group the powers of ten: $10^{-8}\times10^{-2}\times10^{8} = 10^{-2}$. \[ P = \frac{4}{7}\times 5.67 \times 175 \times 10^{-2} \]

Step 5: Finish the arithmetic.
First $\dfrac{4}{7}\times 175 = 4\times 25 = 100$. Then \[ P = 100 \times 5.67 \times 10^{-2} = 5.67\ \text{W}\times 4 = 22.68\ \text{W} \]

Step 6: Conclusion.
The ball loses heat at the rate of 22.68 W. \[ \boxed{22.68\ \text{W}} \]