Q.34 If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Best answer
Answer: \(\dfrac{400}{11}\)°C \(\approx 36.36\)°C
Let \(\theta\) be the temperature of the body at time \(t\) minutes. By Newton’s law of cooling,
\(\dfrac{d\theta}{dt} = -k(\theta - 25)\), \(k \gt 0\) \(\Rightarrow \dfrac{d\theta}{\theta - 25} = -k\,dt\).
Integrating: \(\log(\theta - 25) = -kt + c\).
At \(t = 0\), \(\theta = 80\): \(c = \log 55\). So \(\log\dfrac{\theta - 25}{55} = -kt\), i.e. \(\theta - 25 = 55e^{-kt}\).
At \(t = 30\), \(\theta = 50\): \(25 = 55e^{-30k} \Rightarrow e^{-30k} = \dfrac{5}{11}\).
At \(t = 60\): \(\theta - 25 = 55\left(e^{-30k}\right)^2 = 55\cdot\dfrac{25}{121} = \dfrac{125}{11}\).
\(\theta = 25 + \dfrac{125}{11} = \dfrac{400}{11} \approx 36.36\)°C.
The temperature of the body after 1 hour is about 36.4°C.
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