Newton's Law of Cooling

Problems from IIT JEE

Problem (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)=-\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}

Problem (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}