A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

Heat and Thermodynamics PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q58 with Solution

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Correct answer: A

Use Newton’s law of cooling

For excess temperature over surroundings, $\theta = T - T_s$ and Newton’s law gives $\frac{d\theta}{dt} = -k\theta$ whose solution is $\theta = \theta_0 e^{-kt}.$

So the time taken to cool from temperature $T_1$ to $T_2$ in surroundings at $T_s$ is $t = \frac{1}{k}\ln\left(\frac{T_1-T_s}{T_2-T_s}\right).$

First cooling process: from $90^\circ$ C to $80^\circ$ C

Room temperature: $T_s = 20^\circ \text{C}$

So excess temperatures are $90-20=70, \qquad 80-20=60.$

Given time is $t$ , hence

= \frac{1}{k}\ln\left(\frac{7}{6}\right).$$ --- 3. **Second cooling process: from $80^\circ$C to $60^\circ$C** Excess temperatures are $$80-20=60, \qquad 60-20=40.$$ Let required time be $t'$. Then $$t' = \frac{1}{k}\ln\left(\frac{60}{40}\right) = \frac{1}{k}\ln\left(\frac{3}{2}\right).$$ --- 4. **Find ratio $\dfrac{t'}{t}$** $$\frac{t'}{t} = \frac{\ln(3/2)}{\ln(7/6)}.$$ Now, $$\ln\left(\frac{3}{2}\right) \approx 0.4055, \qquad \ln\left(\frac{7}{6}\right) \approx 0.1542.$$ Thus, $$\frac{t'}{t} \approx \frac{0.4055}{0.1542} \approx 2.63 \approx \frac{13}{5}.$$ Therefore, $$t' \approx \frac{13}{5}t.$$ --- 5. **Check options** - A: $\frac{13}{5}t$ ✅ - B: $\frac{10}{13}t$ ❌ - C: $\frac{5}{13}t$ ❌ - D: $\frac{13}{10}t$ ❌ Hence the correct option is **A**.

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View written solutionFree

Correct answer: A

Use Newton’s law of cooling

For excess temperature over surroundings, $\theta = T - T_s$ and Newton’s law gives $\frac{d\theta}{dt} = -k\theta$ whose solution is $\theta = \theta_0 e^{-kt}.$

So the time taken to cool from temperature $T_1$ to $T_2$ in surroundings at $T_s$ is $t = \frac{1}{k}\ln\left(\frac{T_1-T_s}{T_2-T_s}\right).$

First cooling process: from $90^\circ$ C to $80^\circ$ C

Room temperature: $T_s = 20^\circ \text{C}$

So excess temperatures are $90-20=70, \qquad 80-20=60.$

Given time is $t$ , hence

= \frac{1}{k}\ln\left(\frac{7}{6}\right).$$ --- 3. **Second cooling process: from $80^\circ$C to $60^\circ$C** Excess temperatures are $$80-20=60, \qquad 60-20=40.$$ Let required time be $t'$. Then $$t' = \frac{1}{k}\ln\left(\frac{60}{40}\right) = \frac{1}{k}\ln\left(\frac{3}{2}\right).$$ --- 4. **Find ratio $\dfrac{t'}{t}$** $$\frac{t'}{t} = \frac{\ln(3/2)}{\ln(7/6)}.$$ Now, $$\ln\left(\frac{3}{2}\right) \approx 0.4055, \qquad \ln\left(\frac{7}{6}\right) \approx 0.1542.$$ Thus, $$\frac{t'}{t} \approx \frac{0.4055}{0.1542} \approx 2.63 \approx \frac{13}{5}.$$ Therefore, $$t' \approx \frac{13}{5}t.$$ --- 5. **Check options** - A: $\frac{13}{5}t$ ✅ - B: $\frac{10}{13}t$ ❌ - C: $\frac{5}{13}t$ ❌ - D: $\frac{13}{10}t$ ❌ Hence the correct option is **A**.

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