Question

A body cools from a temperature $3\mathrm{T}$ to $2\mathrm{T}$ in $10$ minutes. The room temperature is $\mathrm{T}$. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next $10$ minutes will be:

• A. $\frac{3}{2}T$
• B. $\frac{7}{4}T$
• C. $\frac{4}{3}T$
• D. $2$$T$

Answer 1

The correct option is A

$\frac{3}{2}T$

Step 1. Given data

The initial temperature $T_1=3\mathrm{T}$

Final temperature $T_2=2\mathrm{T}$

Time $\mathrm{t}=10\min$

Step 2. Formula to be used

Newton's law of cooling:

Newton’s law of cooling the rate of change of temperature of a body through radiation is directly proportional to the difference in the temperature of the body and the surrounding.
Newton’s law of cooling is,

$\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}\left({\mathrm{T}}_{\mathrm{t}}-{\mathrm{T}}_{\mathrm{s}}\right)$

Here, $\mathrm{k}$ is Newton's cooling constant, ${\mathrm{T}}_{\mathrm{t}}$ is temperature of the body at any time $\mathrm{t}$, and ${\mathrm{T}}_{\mathrm{s}}$ temperature of the surrounding.

Step 3. Find the temperature of the body for first $10mins$.

On simplifying the Newton's law of cooling, we get

$\ln \left(\frac{T_1-T_s}{T_2-T_s}\right)=kt$

For the first $10mins$,

$$\begin{gathered} \ln \left(\frac{3T-T}{2T-T}\right)=k\left(10\right) \\ \Rightarrow \ln \left(\frac{2T}{T}\right)=10k \\ \Rightarrow \ln 2=10k-------\left(1\right) \end{gathered}$$

Step 4. Find the temperature of the body for next $10mins$.

Let the temperature at the end of the next $10mins$ be $T\text{'}$

$$\begin{gathered} \ln \left(\frac{2T-T}{T\text{'}-T}\right)=k\left(10\right) \\ \Rightarrow \ln \left(\frac{T}{T\text{'}-T}\right)=10k-------\left(2\right) \end{gathered}$$

Divide equation $\left(1\right)$ and $\left(2\right)$, we get,

$\frac{\ln 2}{\ln \left({\displaystyle \frac{\mathrm{T}}{\mathrm{T}\text{'}-\mathrm{T}}}\right)}=\frac{10\mathrm{k}}{10\mathrm{k}}$

$\frac{\ln 2}{\ln \left({\displaystyle \frac{\mathrm{T}}{\mathrm{T}\text{'}-\mathrm{T}}}\right)}=1$

$\ln \left(\frac{\mathrm{T}}{\mathrm{T}\text{'}-\mathrm{T}}\right)=\ln 2$

Cancelling common log, we get,

$\frac{\mathrm{T}}{\mathrm{T}\text{'}-\mathrm{T}}=2$

$\Rightarrow \mathrm{T}=2\left(\mathrm{T}\text{'}-\mathrm{T}\right)$

$\Rightarrow \mathrm{T}=2\mathrm{T}\text{'}-2\mathrm{T}$

Subtract $\mathrm{T}$ from both sides, we get,

$\mathrm{T}-\mathrm{T}=2\mathrm{T}\text{'}-2\mathrm{T}-\mathrm{T}$

$0=2\mathrm{T}\text{'}-3\mathrm{T}$

$3\mathrm{T}=2\mathrm{T}\text{'}$

$\frac{3}{2}\mathrm{T}=\mathrm{T}\text{'}$

Therefore, the temperature of the body after next $10$ mins is $\frac{3}{2}\mathrm{T}$.

Hence, option C is the correct answer.