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MCQMediumJEE 2024Separation of Variables

JEE Mathematics 2024 Question with Solution

The temperature T(t)T(t) of a body at time t=0t = 0 is 160∘F160^\circ F. If T(15)=120∘FT(15) = 120^\circ F, then T(45)T(45) is:

  • A

    85∘F85^\circ F

  • B

    95∘F95^\circ F

  • C

    90∘F90^\circ F

  • D

    80∘F80^\circ F

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: dTdt=−K(T−80)\frac{dT}{dt} = -K(T-80), T(0)=160∘FT(0)=160^\circ F, and T(15)=120∘FT(15)=120^\circ F.

Find: T(45)T(45).

Separate the variables:

dTT−80=−K dt\frac{dT}{T-80} = -K\,dt

Integrate both sides:

∫dTT−80=∫−K dt\int \frac{dT}{T-80} = \int -K\,dt

So,

ln⁡∣T−80∣=−Kt+C\ln|T-80| = -Kt + C

Exponentiating,

T−80=C′e−KtT-80 = C'e^{-Kt}

Hence,

T(t)=80+C′e−KtT(t)=80+C'e^{-Kt}

Using T(0)=160T(0)=160,

160=80+C′160=80+C'

Therefore,

C′=80C'=80

So the temperature function is

T(t)=80+80e−KtT(t)=80+80e^{-Kt}

Now use T(15)=120T(15)=120:

120=80+80e−15K120=80+80e^{-15K} 40=80e−15K40=80e^{-15K} e−15K=12e^{-15K}=\frac{1}{2}

Thus,

K=ln⁡215K=\frac{\ln 2}{15}

Substitute t=45t=45:

T(45)=80+80e−45⋅ln⁡215T(45)=80+80e^{-45\cdot \frac{\ln 2}{15}} =80+80e−3ln⁡2=80+80e^{-3\ln 2} =80+80(12)3=80+80\left(\frac{1}{2}\right)^3 =80+80⋅18=80+80\cdot \frac{1}{8} =80+10=90=80+10=90

Therefore, the temperature at t=45t=45 is 90∘F90^\circ F. The correct option is C.

Use the ratio directly

Given: dTdt=−K(T−80)\frac{dT}{dt} = -K(T-80) with surrounding temperature 80∘F80^\circ F.

Find: T(45)T(45).

For Newton's law of cooling, the excess temperature T−80T-80 decays exponentially. From t=0t=0 to t=15t=15,

T(0)−80=160−80=80T(0)-80 = 160-80=80

and

T(15)−80=120−80=40T(15)-80 = 120-80=40

So in 1515 minutes, the excess temperature becomes half.

The same factor applies in each equal interval of 1515 minutes. Therefore after 4545 minutes, which is three such intervals,

T(45)−80=80(12)3=10T(45)-80 = 80\left(\frac{1}{2}\right)^3 = 10

Hence,

T(45)=80+10=90T(45)=80+10=90

Therefore, the temperature is 90∘F90^\circ F. The correct option is C.

Common mistakes

  • Using exponential decay on TT directly instead of on T−80T-80. The differential equation is based on the difference from the ambient temperature 80∘F80^\circ F, so the correct quantity to track is T−80T-80, not TT itself.

  • Forgetting to use the condition T(15)=120∘FT(15)=120^\circ F to determine KK. Without finding the decay constant from the second condition, T(45)T(45) cannot be evaluated correctly.

  • Assuming the temperature drops by the same number of degrees every 1515 minutes. The process is exponential, so the excess temperature is multiplied by a constant factor, not decreased by a fixed amount.

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