Question

A body cools down from ${{45}^{0}}C$ to ${{40}^{0}}C$ in 5 minutes and to ${{35}^{0}}C$ in another 8 minutes. Find the temperature of the surroundings.

Answer 1

We are given a body which cools down over a period of time. The intervals of time are also given. We can use Newton's law of cooling in this situation to get the temperature details about the surroundings where the body is kept.

Complete step by step answer:
We know that any body which has a temperature higher than its surroundings has a tendency to cool down and attain equilibrium. This process of radiation of heat to the surrounding is dependent on the temperature of the environment. A greater difference in temperature enables quicker cooling of the same body. This rate of cooling is defined by Newton’s law of cooling. It states that the rate of cooling of a body is directly proportional to the difference in temperature of the body to the surrounding.
Mathematically, it is given as –
$-\dfrac{dT}{dt}=K(T-{{T}_{s}})$
Where, T is the average temperature of the body,
K is the cooling constant,
${{T}_{s}}$ is the temperature of the surrounding.
We can use this relation in this situation to find the temperature of the surrounding as –

& -\dfrac{dT}{dt}=K(T-{{T}_{s}}) \\\ & \text{given, } \\\ & dT={{5}^{0}}C \\\ & \text{dt=5minutes}\times 60=300\text{seconds} \\\ & T={{(\dfrac{45+40}{2})}^{0}}C={{\dfrac{85}{2}}^{0}}C \\\ & \Rightarrow -\dfrac{5}{300}=K(\dfrac{85}{2}-{{T}_{s}})\text{ ---(1)} \\\ \end{aligned}$$ Now, at the second instance again after 8 minutes – $$\begin{aligned} & \text{given, } \\\ & dT={{5}^{0}}C \\\ & \text{dt=8minutes}\times 60=480\text{seconds} \\\ \end{aligned}$$ $$\begin{aligned} & T={{(\dfrac{40+35}{2})}^{0}}C={{\dfrac{75}{2}}^{0}}C \\\ & \Rightarrow -\dfrac{5}{480}=K(\dfrac{75}{2}-{{T}_{s}})\text{ ---(2)} \\\ \end{aligned}$$ Now, we can equate (1) and (2) to get the temperature of the surroundings easily. It is given by – $$\begin{aligned} & -\dfrac{5}{480}=K(\dfrac{85}{2}-{{T}_{s}})\text{ --(2)} \\\ & -\dfrac{5}{300}=K(\dfrac{75}{2}-{{T}_{s}})\text{ --(1) } \\\ & \Rightarrow \dfrac{5}{8}=\dfrac{(\dfrac{75}{2}-{{T}_{s}})}{(\dfrac{85}{2}-{{T}_{s}})} \\\ & \Rightarrow 5(\dfrac{85}{2}-{{T}_{s}})=8(\dfrac{75}{2}-{{T}_{s}}) \\\ & \Rightarrow 3{{T}_{s}}=(\dfrac{600-425}{2}) \\\ & \therefore {{T}_{s}}={{29.16}^{0}}C \\\ \end{aligned}$$ So, we get the temperature of the surroundings of the body to be $${{29.16}^{0}}C$$. This is the required solution for this. **Note:** We should take care not to confuse with the given timings. In this question, the second time is 8 minutes and it should be greater than the first time the body took to decrease its temperature by an equal amount. The rate of cooling decreases with the temperature getting lower, i.e., it takes longer.