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Question: Certain quantities of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5...

Certain quantities of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is ;
A. 45℃
B. 20℃
C. 42℃
C. 10℃

Explanation

Solution

To solve this problem, using the formula for Newton’s Law of cooling. Firstly, consider the case when the water cools from 70℃ to 60℃. Substitute these values in the above mentioned formula. Then, consider the second case when the water cools from 60℃ to 54℃. Substitute these values in the formula. Divide both these obtained equations and solve them. This will give the temperature of the surroundings.

Formula used: dTdt=−K(T−Ts)\dfrac {dT}{dt}= -K ({T}- {T}_{s})

Complete step by step answer:
According to Newton's law of cooling,
dTdt=−K(T−Ts)\dfrac {dT}{dt}= -K ({T}- {T}_{s}) ...(1)
Where, K is the cooling constant
T0{T}_{0} is equal to T1+T02\dfrac{T_1+T_0}{2}
Ts{T}_{s} is the temperature of the surrounding environment
When the temperature cools from 70°C to 60°C then equation. (1) can be written as,
70−605=−K(70+602−Ts)\dfrac {70-60}{5}= -K( \dfrac {70+60}{2}- {T}_{s})
⇒105=−K(65−Ts)\Rightarrow \dfrac {10}{5}= -K(65-{T}_{s})
⇒2=−K(65−Ts)\Rightarrow 2= - K(65-{T}_{s}) ...(2)
Similarly, when the temperature cools from 60°C to 54°C then equation. (1) can be written as,
60−545=−K(60+542−Ts)\dfrac {60-54}{5}= -K( \dfrac {60+54}{2}- {T}_{s})
⇒65=−K(57−Ts)\Rightarrow \dfrac {6}{5}= -K(57-{T}_{s}) ...(3)
Dividing equation. (2) by (3) we get,
53=65−Ts57−Ts\dfrac {5}{3}= \dfrac {65- {T}_{s}}{57-{T}_{s}}
⇒5(57−Ts)=3(65−Ts)\Rightarrow 5(57- {T}_{s})= 3(65- {T}_{s})
⇒285−5Ts=195−3Ts\Rightarrow 285- 5{T}_{s}= 195- 3{T}_{s}
⇒5Ts−3Ts=285−195\Rightarrow 5{T}_{s}-3{T}_{s}= 285-195
⇒2Ts=90\Rightarrow 2{T}_{s}= 90
⇒Ts=45℃\Rightarrow {T}_{s}= 45℃
Hence, the temperature of the surroundings is 45℃.

So, the correct answer is “Option A”.

Note: The rate at which the body changes its temperature is proportional to the difference between the temperature of the object and its environment. This difference is quite small. The rate of cooling of an object also depends upon the cooling constant K. The above mentioned formula can be used for any temperature and environment. Students must remember that Newton’s law of cooling can be applied to both cooling and warming.