Question

A hot body at temperature $T$ losses heat to the surrounding temperature $T_{S}$ by radiation. If the difference in temperature is small then, the rate of loss of heat by the hot body is proportional to

• A. $(T-T_{S})$
• B. $\left(T-T_{S}\right)^{2}$
• C. $\left(T-T_{S}\right)^{1/2}$
• D. $\left(T-T_{S}\right)^{4}$

Answer 1

Answer: (A)

$(T-T_{S})$

Answer 2

By Stefan's law, $\frac{d T}{d t}=\frac{A \varepsilon \sigma}{m c}\left[T^{4}-T_{0}^{4}\right]$ When the temperature difference between the body and its surrounding is not very large ie, $T=T_{0}=\Delta T$ then $T^{4}-T_{0}^{4}$ maybe approximated as $4 T_{0}^{3} \Delta T .$ Hence, $\frac{d T}{d t}=\frac{A \varepsilon \sigma}{m c} 4 T_{0}^{3} \Delta T$ $\Rightarrow \frac{d T}{d t} \propto \Delta T$ ie, if the temperature of the body is not very different from surrounding, rate of cooling is proportional to temperature difference between the body and its surrounding. This law is called Newton's law of cooling.