# Newton's Law of Cooling

Newton's law of cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings i.e., \begin{align} \mathrm{d}Q/\mathrm{d}t=-b_1 (T-T_s), \end{align} where $T$ is the temperature of the body, $T_s$ is the temperature of the surrounding and $b_1$ is heat transfer coefficient (constant). The negative sign indicates that the body loses heat if $T>T_s$.

If the difference in temperature between the body and its surroundings is not large then Newton’s law of cooling holds good irrespective of the mode of heat transfer - conductive, convective or radiative. For radiative heat transfer, Newton's law of cooling can be derived from Stefan-Boltzmann law.

The rate of cooling is defined as decrease in temperature per unit time ($\mathrm{d}T/\mathrm{d}t$). Substitute $\mathrm{d}Q=ms\mathrm{d}T$ in above equation to get a useful form of Newton’s law of cooling \begin{align} \mathrm{d}T/\mathrm{d}t=-b (T-T_0). \end{align}

This differential equation is solved by using initial conditions. If $T_0$ is the initial temperature of the body then its temperature at time $t$ is given by \begin{align} T=T_s+(T_0-T_s)e^{-bt}.\nonumber \end{align} Newton’s Law of Cooling. The rate of cooling (magnitude of the slope) decreases as time progresses.

Following points are worth noting

1. The rate of cooling is high in the beginning but it decreases as time progresses. In figure, it takes 5 min for the temperature to decreases from 90℃ to 70℃ but it takes 7.5 min to decrease it from 70℃ to 50℃.
2. The time required for body temperature to reach the surrounding temperature is infinite.
3. The rate of cooling can be increased by increasing the heat transfer coefficient.

## Solved Problems on Newton's Law of Cooling

### Example Problem 1

The temperature of a body falls from 90℃ to 70℃ in 5 minutes when placed in a surrounding of constant temperature 20℃. Find the time taken for the body to become 50℃.

Solution: Integrate the differential equation of Newton's law of cooling from time $t=0$ to $t=5$ min to get \begin{align} \int_{90}^{70}\frac{\mathrm{d}T}{T-20}=-b\int_0^5 \mathrm{d}t \nonumber\\ \ln\left(\frac{70-20}{90-20}\right)=-5b,\nonumber \end{align} which gives $b=(1/5)\ln(7/5)$. Now, repeat the same for time interval $t=5$ min to $t=\tau$ in which temperature decreases from 70℃ to 50℃. \begin{align} \int_{70}^{50}\frac{\mathrm{d}T}{T-20}=-b\int_5^\tau \mathrm{d}t \nonumber\\ \ln\left(\frac{50-20}{70-20}\right)=-b(5-\tau).\nonumber \end{align} Substitute $b$ and simplify to get $\tau=12.6$ min. The time taken to cool from 70℃ to 50℃ is $12.6-5=7.6$ min.

### Example Problem 2

The temperature of a body falls from 40℃ to 36℃ in 5 minutes when placed in a surrounding of constant temperature 16℃. Find the time taken for the temperature of the body to become 32℃.

Solution: We solve this problem by an approximation method that makes use of mean temperature. The approximation errors are small because temperature differences are not large.

The mean temperature of the body as it cools from 40℃ to 36℃ is $T_m=(40+36)/2=38$℃. Newton's law of cooling can be written as \begin{align} \Delta T/\Delta t=-b (T_m-T_s), \nonumber \\ (36-40)/5=-b(38-16), \end{align} which gives $b=0.8/22$/min. Note that $\Delta T/\Delta t$ is negative.

Let the time taken for the temperature to become 32℃ be $t$. During this period, $T_m=(36+32)/2=34$℃. Substitute in above equation to get \begin{align} (32-36)/t=-b(34-16), \nonumber \\ -4/t=-(0.8/22)(18), \nonumber \end{align} which gives $t=6.1$ min.

### Problems from IIT JEE 1982

A solid sphere of copper of radius $R$ and a hollow sphere of the same material of inner radius $r$ and outer radius $R$ are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster?

Solution: Both the spheres have same surface area $A$, same emissivity $e$, same temperature $T$, and same ambient temperature $T_0$. The net rate of heat radiation by both the spheres is equal to \begin{align} \mathrm{d}Q/\mathrm{d}t=\sigma e A (T^4-T_0^4).\nonumber \end{align} The rate of heat radiation is equal to the rate of heat loss due to decrease in temperature i.e., \begin{align} \mathrm{d}Q/\mathrm{d}t=mS(-\mathrm{d}T/\mathrm{d}t),\nonumber \end{align} where $m$ is the mass of the sphere and $S$ is the specific heat. From above equations, the rate of cooling is given by \begin{align} -\frac{\mathrm{d}T}{\mathrm{d}t}=\frac{\sigma e A(T^4-T_0^4)}{mS}.\nonumber \end{align} Since mass of the hollow sphere is less than that of the solid sphere, the rate of cooling of the hollow sphere is more than that of the solid sphere.

### Problems from IIT JEE 1991

A solid copper sphere (density $\rho$ and specific heat $c$) of radius $r$ at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required for the temperature of the sphere to drop to 100 K is ___________?

Solution: Stefan's law gives the rate of energy radiated by the sphere as \begin{align} {\mathrm{d}Q}/{\mathrm{d}t}=\sigma A T^4=4\pi\sigma r^2 T^4. \nonumber \end{align} The temperature of the sphere reduces due to energy loss. The rate of energy loss is related to the rate of temperature decrease by \begin{align} {\mathrm{d}Q}/{\mathrm{d}t}&=-mc\left({\mathrm{d}T}/{\mathrm{d}t}\right) \nonumber\\ &=-({4}/{3})\pi r^3 \rho c \left({\mathrm{d}T}/{\mathrm{d}t}\right). \nonumber \end{align} Eliminate ${\mathrm{d}Q}/{\mathrm{d}t}$ from above equations to get \begin{align} \rho r c\left( {\mathrm{d}T}/{\mathrm{d}t}\right)=-{3\sigma} T^4. \nonumber \end{align} Integrate above equation \begin{align} \rho r c \int_{T_i}^{T_f}T^{-4}\,\mathrm{d}T=-{3\sigma}\int_0^t\mathrm{d}t,\nonumber \end{align} to get \begin{align} t&=\frac{\rho rc}{9\sigma}\left[\frac{1}{T_f^3}-\frac{1}{T_i^3}\right] \nonumber\\ &=\frac{\rho r c}{9(5.67\times{10}^{-8})}\left[\frac{1}{100^3}-\frac{1}{200^3}\right] \nonumber \\ &=1.71\rho r c.\nonumber \end{align}

### Problems from IIT JEE 1998

A solid body X of heat capacity $C$ is kept in an atmosphere whose temperature is $T_A=300\;\mathrm{K}$. At time $t=0$, the temperature of X is $T_0=400\;\mathrm{K}$. It cools according to Newton's law of cooling. At time $t_1$ its temperature is found to be 350 K. At this time $(t_1)$ the body X is connected to a large body Y at atmospheric temperature $T_A$ through a conducting rod of length $L$, cross-sectional area $A$ and thermal conductivity $K$. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area $A$ of the connecting rod is small compared to the surface area of X. Find the temperature of X at time $t=3t_1$.

Solution: In first case, the body X loses thermal energy due to radiation. Newton's law of cooling for a body at temperature $T$ kept in an atmosphere at temperature $T_A$ gives rate of cooling, \begin{align} {\mathrm{d}T}/{\mathrm{d}t}=-k(T-T_A). \end{align} Integrate from $t=0$ to $t=t_1$, \begin{align} \int_{T_0}^{T_1}\frac{\mathrm{d}T}{T-T_A}=-k\int_0^{t_1}\mathrm{d}t, \nonumber \end{align} to get, \begin{align} kt_1 &=-\ln\left(\frac{T_1-T_A}{T_0-T_A}\right)\nonumber\\ &=-\ln\left(\frac{350-300}{400-300}\right)=\ln 2. \end{align} The rate of heat loss is related to the rate of temperature change by, \begin{align} {\mathrm{d}Q_\text{r}}/{\mathrm{d}t}=-mS\left(\mathrm{d}T/\mathrm{d}t\right)= -C\left(\mathrm{d}T/\mathrm{d}t\right). \end{align} Substitute $\mathrm{d}T/\mathrm{d}t$ and $k$ from to get, \begin{align} {\mathrm{d}Q_\text{r}}/{\mathrm{d}t}=Ck(T-T_A)={C}(T-T_A)\ln 2/{t_1}. \end{align}

In second case, heat is lost to the atmosphere by radiation and to the body $Y$ by conduction. The rate of heat lost by $X$ due to conduction is, \begin{align} {\mathrm{d}Q_\text{c}}/{\mathrm{d}t}={KA(T-T_A)}/{L}. \end{align} These equations give total rate of heat lost by $X$ as, \begin{align} \mathrm{d}Q_t/\mathrm{d}t&={\mathrm{d}Q_\text{r}}/{\mathrm{d}t}+{\mathrm{d}Q_\text{c}}/{\mathrm{d}t}\nonumber\\ &=C(T-T_A){\ln 2}/{t_1}+{KA(T-T_A)}/{L}\nonumber\\ &=C\left[{\ln 2}/{t_1}+{KA}/{(LC)}\right](T-T_A). \end{align} which is related to the rate of temperature change of $X$ by, \begin{align} \mathrm{d}Q_t/\mathrm{d}t=-C{\mathrm{d}T}/{\mathrm{d}t}. \end{align} From these equations, \begin{align} \mathrm{d}T/\mathrm{d}t= -\left[{\ln 2}/{t_1}+{KA}/{(LC)}\right](T-T_A) \end{align} Integrate from $t=t_1$ to $t=3t_1$, \begin{align} \int_{T_1}^{T_2}\frac{\mathrm{d}T}{T-T_A}=-\left(\frac{\ln 2}{t_1}+\frac{KA}{LC}\right)\int_{t_1}^{3t_1}\mathrm{d}t,\nonumber \end{align} to get, \begin{align} \ln\left(\frac{T_2-T_A}{T_1-T_A}\right)=-2\ln2-\frac{2KAt_1}{LC}. \end{align} Substitute the values in above equation and simplify to get, \begin{align} T_2=300+12.5 e^{-\frac{2KAt_1}{LC}}.\nonumber \end{align}

### Problems from IIT JEE 1995

Two metallic spheres $S_1$ and $S_2$ are made of the same material and have got identical surface finish. The mass of $S_1$ is thrice that of $S_2$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_1$ to that of $S_2$ is,

1. $\frac{1}{3}$
2. $\frac{1}{\sqrt{3}}$
3. $\sqrt{3}$
4. $\left(\frac{1}{3}\right)^{1/3}$

Solution: Let $r_1$ and $r_2$ be the radii of $S_1$ and $S_2$ and $\rho$ be density of the material. The masses of $S_1$ and $S_2$ are, \begin{align} &m_1=({4}/{3})\pi r_1^3 \rho, &&m_2=({4}/{3})\pi r_2^3 \rho.\nonumber \end{align} Since $m_1=3m_2$, we get $r_1=3^{1/3}r_2$.

Let $T$ be temperature of two spheres and $T_0$ be room temperature. The rate of radiation heat loss by $S_1$ and $S_2$ are, \begin{align} \label{qoa:eqn:1} {\mathrm{d}Q_1}/{\mathrm{d}t}=4\pi \sigma e r_1^2 (T^4-T_0^4), \\ \label{qoa:eqn:2} {\mathrm{d}Q_2}/{\mathrm{d}t}=4\pi \sigma e r_2^2 (T^4-T_0^4). \end{align} The temperature of the sphere reduces due to radiation heat loss. The rate of temperature change for $S_1$ and $S_2$ are given by, \begin{align} \label{qoa:eqn:3} &{\mathrm{d}Q_1}/{\mathrm{d}t}=-m_1S({\mathrm{d}T_1}/{\mathrm{d}t}),\\ \label{qoa:eqn:4} &{\mathrm{d}Q_2}/{\mathrm{d}t}=-m_2S({\mathrm{d}T_2}/{\mathrm{d}t}). \end{align} Eliminate ${\mathrm{d}Q_1}/{\mathrm{d}t}$ and ${\mathrm{d}Q_2}/{\mathrm{d}t}$ from to get, \begin{align} \label{qoa:eqn:5} & {\mathrm{d}T_1}/{\mathrm{d}t}=-{3 \sigma e (T^4-T_0^4)}/{(r_1 \rho S)},\\ \label{qoa:eqn:6} & {\mathrm{d}T_2}/{\mathrm{d}t}=-{3 \sigma e (T^4-T_0^4)}/{(r_2 \rho S)}. \end{align} Divide above equations to get the ratio of rate of cooling as, \begin{align} \frac{\mathrm{d}T_1/\mathrm{d}t}{\mathrm{d}T_2/\mathrm{d}t}=\frac{r_2}{r_1}=\left(\frac{1}{3}\right)^{1/3}.\nonumber \end{align}

## Questions on Newton's Law of Cooling

Question 1: A solid at temperature $T_1$ is kept in an evacuated chamber at temperature $T_2>T_1$. The rate of increase of temperature of the body is proportional to

A. $T_2-T_1$
B. $T_2^2-T_1^2$
C. $T_2^3-T_1^3$
D. $T_2^4-T_1^4$

Question 2: In a murder investigation, a corpse was found by a detective at exactly 8 P.M. Being alert, the detective also measured the body temperature and found it to be 70°F. Two hours later, the detective measured the body temperature again and found it to be 60°F. If the room temperature is 50°F, and assuming that the body temperature of the person before death was 98.6° F, at what time did the murder occur?

A. 7:30 PM
B. 6:30 PM
C. 5:30 PM
D. 4:30 PM