Variable Separable
A cup of coffee cools according to Newton's law so that \frac{dT}{dt} = -k(T - T_s); identify the correct form of the temperature as a function of time.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- variable separable
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
T=Ts+(T0−Ts)e−kt
Newton's law of cooling states that the rate of temperature change is proportional to the difference between the object's temperature and the surrounding temperature T_s, giving a separable first-order equation. The key idea is to separate the temperature difference from time, treating T - T_s as the relevant variable. Let u = T - T_s, so \frac{du}{dt} = \frac{dT}{dt} = -ku. This separates to \frac{du}{u} = -k,dt, and integrating gives \ln|u| = -kt + C_1, hence u = Ce^{-kt}. Restoring u = T - T_s gives T - T_s = Ce^{-kt}. Applying the initial temperature T_0 at t = 0 gives C = T_0 - T_s, so T = T_s + (T_0 - T_s)e^{-kt}. Option with e^{kt} would describe heating without bound, contradicting cooling. Option T = T_0 e^{-kt} ignores the ambient temperature entirely. Option with a minus sign before (T_0 - T_s) gives the wrong initial value at t = 0. This is the standard JEE Advanced cooling application. As a final plausibility check, as t \to \infty the exponential vanishes and T \to T_s, correctly modeling that the coffee eventually reaches room temperature.
This medium difficulty mathematics question is from the chapter differential equations, covering the topic of variable separable. It appeared in the 2025 exam.
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