Question

A body cools down from 50°C to 45°C in 5 mintues and to 40°C in another 8 minutes. Find the temperature of the surrounding.

Answer 1

Let the temperature of the surroundings be T₀°C.

Case 1:
Initial temperature of the body = 50°C
Final temperature of the body = 45°C
Average temperature = 47.5 °C

Difference in the temperatures of the body and its surrounding = (47.5 − T)°C
Rate of fall of temperature = $\frac{∆T}{t}=\frac{5}{5}=1^{°}\mathrm{C}/\min$

By Newton's law of cooling,

$$\begin{gathered} \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}\left[T_{\mathrm{avg}}-{\mathrm{T}}_0\right] \\ 1=-\mathrm{K}\left[47.5-{\mathrm{T}}_0\right]...\left(1\right) \end{gathered}$$

Case 2:
Initial temperature of the body = 45°C
Final temperature of the body = 40°C
Average temperature = 42.5 °C

Difference in the temperatures of the body and its surrounding = (42.5 − T₀)°C
Rate of fall of temperature = $\frac{∆T}{t}=\frac{5}{8}={\frac{5}{8}}^{°}\mathrm{C}/\min$
From Newton^,s law of cooling,
$$\begin{gathered} \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}\left[T_{\mathrm{avg}}-{\mathrm{T}}_0\right] \\ 0.625=-\mathrm{K}\left[42.5-{\mathrm{T}}_0\right]...\left(2\right) \end{gathered}$$

Dividing (1) by (2),

$$\begin{gathered} \frac{1}{0.625}=\frac{47.5-T_0}{42.5-T_0} \\ \Rightarrow 42.5-T_0=29.68-0.625T \\ \Rightarrow T_0={34}^{°}C \end{gathered}$$