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MCQMediumJEE 2023Heat Transfer (Conduction, Convection, Radiation)

JEE Physics 2023 Question with Solution

A bowl filled with very hot soup cools from 98∘C98^\circ C to 86∘C86^\circ C in 22 minutes when the room temperature is 22∘C22^\circ C. How long will it take to cool from 75∘C75^\circ C to 69∘C69^\circ C?

  • A

    22 minutes

  • B

    1.41.4 minutes

  • C

    33 minutes

  • D

    11 minute

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The soup cools from 98∘C98^\circ C to 86∘C86^\circ C in 22 minutes, and the room temperature is 22∘C22^\circ C.

Find: The time required to cool from 75∘C75^\circ C to 69∘C69^\circ C.

According to Newton's Law of Cooling,

ΔQΔt=−k(T−T0)\frac{\Delta Q}{\Delta t} = -k (T - T_0)

where T0T_0 is the room temperature and TT is the average temperature of the body during the time interval.

For the first cooling phase 98∘C→86∘C98^\circ C \to 86^\circ C,

Δt=2 min,Tavg=98+862=92∘C\Delta t = 2 \, \text{min}, \quad T_{\text{avg}} = \frac{98 + 86}{2} = 92^\circ C

Substituting,

k=ΔQΔt=12×(92−22)=702k = \frac{\Delta Q}{\Delta t} = \frac{1}{2} \times (92 - 22) = \frac{70}{2}

For the second cooling phase 75∘C→69∘C75^\circ C \to 69^\circ C,

Tavg=75+692=72∘CT_{\text{avg}} = \frac{75 + 69}{2} = 72^\circ C

Substituting,

Δt=ΔQk=1702×(72−22)=65\Delta t = \frac{\Delta Q}{k} = \frac{1}{\frac{70}{2}} \times (72 - 22) = \frac{6}{5}

Thus,

Δt=1.4 minutes\Delta t = 1.4 \, \text{minutes}

Therefore, the time taken is 1.4 minutes1.4 \, \text{minutes}. The solution concludes that the correct option is C, although the computed value matches option B.

Answer Discrepancy Note

The solution working gives the final value as 1.4 minutes1.4 \, \text{minutes}. Among the listed options, this matches option B. However, the solution states "The Correct Option is C," which is inconsistent with the numerical result. Since the working clearly supports 1.4 minutes1.4 \, \text{minutes}, the defensible answer from the provided material is the option containing that value.

Common mistakes

  • Using the initial temperature difference instead of the average temperature over the interval is incorrect here because the provided solution applies Newton's law using mean temperature for each short interval. Use T1+T22\frac{T_1+T_2}{2} for each cooling range before substituting.

  • Ignoring the room temperature 22∘C22^\circ C leads to a wrong proportionality. The cooling rate depends on the excess temperature T−T0T-T_0, not on the body temperature alone. Always subtract the surrounding temperature first.

  • Choosing option C only because the header says so is risky when the actual calculation gives 1.41.4 minutes. Compare the computed value with the options and note any mismatch explicitly.

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