Heat Transfer
7.0 Newton's laws of cooling
7.0 Newton's laws of cooling
Consider a body of surface area $A$ at an absolute temperature $T$ and the surroundings at ${T_0}$, as discussed before, the net loss of thermal energy can be written as$$\Delta {u_1} = e\sigma A\left( {{T^4} - {T^4}_0} \right)$$If temperature difference is very small, $T = {T_0} + \Delta T$, then$${{T^4} - {T^4}_0 = {{\left( {{T_0} + \Delta T} \right)}^4} - {T_0} \approx 4T_0^3\Delta T = 4T_0^3\left( {T - {T_0}} \right)}$$Thus, $$\begin{equation} \begin{aligned} \Delta {u_1} = 4e\sigma AT_0^3\left( {T - {T_0}} \right) \\ = {b_1}A\left( {T - {T_0}} \right) \\\end{aligned} \end{equation} $$The energy is also lost due to convection.
For small temperature difference, the heat loss per unit time is proportional to area & temperature difference.$$\Delta {u_2} = {b_2}A\left( {T - {T_0}} \right)$$The net heat loss can be written as$$\Delta u = \Delta {u_1} + \Delta {u_2} = \left( {{b_1} + {b_2}} \right)A\left( {T - {T_0}} \right)$$If $m$ is the mass and $s$ is the specific heat of the body,$$\begin{equation} \begin{aligned} \frac{{ - dT}}{{dt}} = \frac{{\Delta u}}{{ms}} = \frac{{\left( {{b_1} + {b_2}} \right)A\left( {T - {T_0}} \right)}}{{ms}} \\ = bA\left( {T - {T_0}} \right) \\ \Rightarrow \frac{{dT}}{{dt}} = - bA\left( {T - {T_0}} \right) \\\end{aligned} \end{equation} $$This is known as Newton's law of cooling.
It states that for small temperature difference, the rate of heat transfer is proportional to the temperature difference between the body and the surrounding.
The minus sign indicates that temperature decreases with time.
We can also write the above equation as,$$\begin{equation} \begin{aligned} \frac{{dT}}{{dt}} = - b\left( {T - {T_0}} \right) \Rightarrow \frac{{dT}}{{T - {T_0}}} = - bdt \\ \Rightarrow bt = \ln \left( {\frac{{{T_1} - {T_0}}}{{{T_2} - {T_0}}}} \right) \\\end{aligned} \end{equation} $$This gives the time taken by a body to cool down from ${{T_1}}$ to ${{T_2}}$.
Question 9. A body cools in $10\ minutes$ from ${60^o}C$ to ${40^o}C$. What will be its temperature after $10\ minutes$? The temperature of the surroundings is ${10^o}C$.
Solution: According to the Newton's law of cooling$$\begin{equation} \begin{aligned} \left( {\frac{{{T_1} - {T_2}}}{t}} \right) = \alpha \left[ {\left( {\frac{{{T_1} + {T_2}}}{2}} \right) - {T_0}} \right] \\ \frac{{60 - 40}}{{10}} = \alpha \left[ {\frac{{60 + 40}}{2} - 10} \right] .....(1) \\\end{aligned} \end{equation} $$Let $T$ be the temperature after next $10\ minutes$. Then, $$\frac{{40 - T}}{{10}} = \alpha \left[ {\frac{{40 + T}}{2} - 10} \right]\quad ....(2)$$Solving (1) and (2) equations, we get $T$ = ${28^o}C$.
Question 10. Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$. The outer surface areas of two bodies are equal and they emit total radiant power at the same rate. The wave length ${\lambda _B}$ corresponding to maximum spectral radiancy in the radiation from $B$ is shifted from the wave length corresponding to maximum spectral radiancy in the radiation from $A$ by $1.00\mu m$. If the temperature of $A$ is $5802\ K$, Find
(a) the temperature of $B$
(b) wave length ${\lambda _B}$.
Solution: (a) According to stefan's law, the power radiated by a body is given by$$u = e\sigma A{T^4}$$Given that$$\begin{equation} \begin{aligned} {u_A} = {u_B},{A_A} = {A_B} \\ so,\quad {e_A}T_A^4 = {e_B}T_B^4 \\ 0.01 \times {\left( {5802} \right)^2} = 0.81{\left( {{T_B}} \right)^4} \\ {T_B} = \left( {\frac{1}{3}} \right)\left( {5802} \right) = 1934\,K \\\end{aligned} \end{equation} $$
(b) According to Wein's displacement law$$\begin{equation} \begin{aligned} {\lambda _A}{T_A} = {\lambda _B}{T_B} \\ {\lambda _B} = \left( {\frac{{5802}}{{1934}}} \right){\lambda _A} \\ {\lambda _B} = 3{\lambda _A}\quad and\quad {\lambda _B} - {\lambda _A} = 1\,\mu m \\ \Rightarrow {\lambda _B} - \left( {\frac{1}{3}} \right){\lambda _A} = 1\,\mu m \\ {\lambda _B} = 1.5\,\mu m \\\end{aligned} \end{equation} $$
Question 12. A point source of heat of power $P$ is placed at the centre of a spherical shell of mean radius $R$. The material of the shell has thermal conductivity $K$. Calculate the thickness of the shell if the temperature difference between the outer and inner surfaces of the shell in steady state is $T$.
Solution: Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure. In, steady state, the rate of flow of heat through shell be$$\begin{equation} \begin{aligned} \frac{{dQ}}{{dt}} = \frac{{\Delta T}}{R} = \frac{{\frac{{\left( { - dT} \right)}}{{dr}}}}{{\left( K \right)\left( {4\pi {r^2}} \right)}}\quad \left( {as\;R = \frac{l}{{KA}}} \right) \\ \frac{{dQ}}{{dt}} = - \left( {4\pi K{r^2}} \right)\frac{{dT}}{{dr}} \\\end{aligned} \end{equation} $$Here, negative sign is used because with increase in $r$, $T$ decreases.$$\begin{equation} \begin{aligned} \int\limits_{{r_1}}^{{r_2}} {\frac{{dr}}{{{r^2}}}} = - \frac{{4\pi k}}{{\left( {\frac{{dQ}}{{dt}}} \right)}}\int\limits_{{T_2}}^{{T_2}} {dT} \\ \frac{{dQ}}{{dt}} = \frac{{4\pi K{r_1}{r_2}\left( {{T_1} - {T_2}} \right)}}{{{r_2} - {r_1}}} \\\end{aligned} \end{equation} $$In steady state, $$\frac{{dQ}}{{dt}} = P,{r_1}{r_2} \approx {R^2},{T_1} - {T_2} = T$$Thickness of shell is, $${r_2} - {r_1} = \frac{{4\pi K{R^2}T}}{P}$$
