Question

The rate at which a body cools is proportional to the difference between the temperature of the body and that of the surrounding air. If a body in air at 25°C will cool from 100° to 75° in one minute, find its temperature at the end of three minutes.

Answer 1

Solution

find its temperature at the end of three minutes.

Given Rok of coording \( \alpha (T_h-T_{min}) \)

\( \implies \frac{dT}{dt}=-k(T-T_h) \)

\( \implies \frac{dT}{(T-T_h)}=-kdt \)

\( \implies \int_{T_0}^{T}\frac{dT}{(T-T_h)}=-k\int_{0}^{t} dt \)

\( \implies \int_{T_0}^{T}\frac{dT}{(T-25)}=-k\int_{0}^{t} dt \)

\( \implies ln\Big(\frac{T-25}{T_0-25}\Big)=-kt \)

\( \bullet\quad At \quad t=1min ,\quad T=75 \)

\( \implies ln\Big(\frac{75-25}{100-25}\Big)=-k(1) \)

\( \implies ln\Big(\frac{50}{75}\Big)=-k \)

\( \implies k=0.4054\quad min^{-1} \)

\( \bullet\quad At \quad t=3min \)

\( \implies ln\Big(\frac{T-25}{75}\Big)=-3(0.4054) \)

\( \implies\frac{T-25}{75} = e^{-3(0.4054)} \)

\( \implies\frac{T-25}{75} = 0.296 \)

Therefore.​​​​​​​ T=47.2°C