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Question: Hot water cools from \(60{}^\circ C\) to \(50{}^\circ C\) in the first 10 minutes and to \(42{}^\cir...

Hot water cools from 60∘C60{}^\circ C to 50∘C50{}^\circ C in the first 10 minutes and to 42∘C42{}^\circ C in the next 10 minutes. The temperature of the surroundings is:
A. 25∘C25{}^\circ C
B. 10∘C10{}^\circ C
C. 15∘C15{}^\circ C
D. 20∘C20{}^\circ C

Explanation

Solution

We could solve the question by using Newton’s law of cooling. We are given two cases, one where the temperature cools from 60∘C60{}^\circ C to 50∘C50{}^\circ C and then from 50∘C50{}^\circ C to 42∘C42{}^\circ C. We could substitute these values in the expression of Newton’s law of cooling and thus form two linear equations in two variables and hence solve them to get the surrounding temperature.
Formula used:
Newton’s law of cooling,
⇒Tt=Ts+(T0−Ts)e−Kt\Rightarrow {{T}_{t}}={{T}_{s}}+\left( {{T}_{0}}-{{T}_{s}} \right){{e}^{-Kt}}

Complete step-by-step solution
We are given some hot water that cools from 60∘C60{}^\circ C to 50∘C50{}^\circ C in the first 10 minutes and then it further cools down to 42∘C42{}^\circ C in the next 10 minutes. With this information, we are asked to find the temperature of the surroundings.
Newton’s law of cooling states that the rate of loss of heat (−dQdt)\left( -\dfrac{dQ}{dt} \right) is directly proportional to the difference of temperature of body and the surroundings (ΔT=(T−Ts))\left( \Delta T=\left( T-{{T}_{s}} \right) \right). But it is to be noted that this law only holds good for small differences in temperature.
−dQdt∝(T−Ts)-\dfrac{dQ}{dt}\propto \left( T-{{T}_{s}} \right)
⇒−dQdt=k(T−Ts)\Rightarrow -\dfrac{dQ}{dt}=k\left( T-{{T}_{s}} \right) ……………………….. (1)
Where,
TT = temperature of the body
Ts{{T}_{s}} = temperature of the surroundings
Also, k is a positive constant that depends upon the area and nature of the surface of the body.
Let,
m = mass of a body
s = specific heat capacity
Ts{{T}_{s}} = temperature of the surroundings
TT = temperature of the body
If the temperature reduces by dTdT in time dtdt, then heat lost is given by,
dQ=msdTdQ=msdT
Rate of loss of heat could be given by,
dQdt=msdTdt\dfrac{dQ}{dt}=ms\dfrac{dT}{dt} ………………………. (2)
From (1) and (2), we have,
⇒−msdTdt=k(T−Ts)\Rightarrow -ms\dfrac{dT}{dt}=k\left( T-{{T}_{s}} \right)
⇒dTT−Ts=−kmsdt=−Kdt\Rightarrow \dfrac{dT}{T-{{T}_{s}}}=-\dfrac{k}{ms}dt=-Kdt ………………………. (3)
Where, K=kmsK=\dfrac{k}{ms}
Let,
T0{{T}_{0}} = temperature of the body at time t=0
Tt{{T}_{t}} = temperature of the body at time t
Integrating (3) on both sides,
∫T0TtdTT−Ts=−K∫0tdt\int\limits_{{{T}_{0}}}^{{{T}_{t}}}{\dfrac{dT}{T-{{T}_{s}}}}=-K\int\limits_{0}^{t}{dt}
We know that, ∫1xdx=ln⁡x\int{\dfrac{1}{x}}dx=\ln x
⇒[ln⁡(T−Ts)]T0Tt=−K[t]0t\Rightarrow \left[ \ln \left( T-{{T}_{s}} \right) \right]_{{{T}_{0}}}^{{{T}_{t}}}=-K\left[ t \right]_{0}^{t}
⇒ln⁡(Tt−Ts)−ln⁡(T0−Ts)=−K(t−0)\Rightarrow \ln \left( {{T}_{t}}-{{T}_{s}} \right)-\ln \left( {{T}_{0}}-{{T}_{s}} \right)=-K\left( t-0 \right)
But,ln⁡a−ln⁡b=ln⁡ab\ln a-\ln b=\ln \dfrac{a}{b}
⇒ln⁡(Tt−TsT0−Ts)=−Kt\Rightarrow \ln \left( \dfrac{{{T}_{t}}-{{T}_{s}}}{{{T}_{0}}-{{T}_{s}}} \right)=-Kt
Taking antilog on both sides,
⇒(Tt−Ts)(T0−Ts)=e−Kt\Rightarrow \dfrac{\left( {{T}_{t}}-{{T}_{s}} \right)}{\left( {{T}_{0}}-{{T}_{s}} \right)}={{e}^{-Kt}}
⇒Tt=Ts+(T0−Ts)e−Kt\Rightarrow {{T}_{t}}={{T}_{s}}+\left( {{T}_{0}}-{{T}_{s}} \right){{e}^{-Kt}}…………………….. (4)
We can now substitute the given values in this equation to find the surrounding temperature.
Case 1: Temperature is decreased from 60∘C60{}^\circ C to 50∘C50{}^\circ C in first 10 minutes
50=Ts+(60−Ts)e−10K50={{T}_{s}}+\left( 60-{{T}_{s}} \right){{e}^{-10K}}
⇒50=Ts(1−e−10K)+60e−10K\Rightarrow 50={{T}_{s}}\left( 1-{{e}^{-10K}} \right)+60{{e}^{-10K}}
Let,
⇒Ts(1−e−10K)=a\Rightarrow {{T}_{s}}\left( 1-{{e}^{-10K}} \right)=a…………………… (5)
⇒e−10K=b\Rightarrow {{e}^{-10K}}=b ………………………. (6)
⇒50=a+60b\Rightarrow 50=a+60b …………………….. (7)
Case 2: Temperature is further decreased from 50∘C50{}^\circ C to 42∘C42{}^\circ C in next 10 minutes
⇒42=Ts+(50−Ts)e−10K\Rightarrow 42={{T}_{s}}+\left( 50-{{T}_{s}} \right){{e}^{-10K}}
⇒42=Ts(1−e−10K)+50e−10K\Rightarrow 42={{T}_{s}}\left( 1-{{e}^{-10K}} \right)+50{{e}^{-10K}}
⇒42=a+50b\Rightarrow 42=a+50b ……………………… (8)
Subtracting (8) from (7),
⇒8=10b\Rightarrow 8=10b
⇒b=45\Rightarrow b=\dfrac{4}{5} …………… (9)
Adding (7) and (8) gives,
⇒92=2a+110b\Rightarrow 92=2a+110b
⇒2a+110×45=92\Rightarrow 2a+110\times \dfrac{4}{5}=92
⇒2a=4\Rightarrow 2a=4
⇒a=2\Rightarrow a=2 ………………. (10)
From (6) and (9),
⇒b=e−10K=45\Rightarrow b={{e}^{-10K}}=\dfrac{4}{5}
From (5) and (10),
⇒a=Ts(1−e−10K)=2\Rightarrow a={{T}_{s}}\left( 1-{{e}^{-10K}} \right)=2
⇒Ts(1−45)=2\Rightarrow {{T}_{s}}\left( 1-\dfrac{4}{5} \right)=2
⇒Ts5=2\Rightarrow \dfrac{{{T}_{s}}}{5}=2
⇒Ts=10∘C\Rightarrow {{T}_{s}}=10{}^\circ C
Therefore, the temperature of the surroundings is 10∘C10{}^\circ C.
Hence, the answer to the question is option B.

Note: Though for conceptual clarity we have derived the expression for Newton’s law of cooling you shouldn’t necessarily do that. You could simply memorize the standard expression and then simply substitute the given values and hence get the answer. In case you are not able to recall the expression, you can derive it as shown above.