
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T, assume that Newton’s law of cooling is applicable. The temperature of the body at the end of the next 10 minutes will be?
A. \[\dfrac{7T}{4}\]
B. \[\dfrac{3T}{2}\]
C. \[\dfrac{4T}{3}\]
D. $T$
Answer
558.6k+ views
Hint:This problem can be solved easily by using Newton's law of cooling as we have to determine the rate of cooling. We are provided a condition in which a body cools down from a certain temperature to another, we need to determine the temperature of the body after 10 minutes. The temperature is given in the form of variables.
Complete step by step answer:
Newton's Law of Cooling, rate of heat transfer is given by
\[\dfrac{dQ}{dt}=kA(T-{{T}_{0}})\]
The surrounding temperature is constant at T. The body takes time 10 minutes to cool from 3T to T. Writing this mathematically, from Newton’s law of cooling,
$\ln \dfrac{3T-T}{2T-T}=kt \\
\Rightarrow \ln 2=10k \\
\therefore k=\dfrac{\ln 2}{10} \\$
Now for next 10, minutes
$\ln \dfrac{2T-T}{T'-T}=kt \\
\Rightarrow \ln \dfrac{2}{T'-T}=\dfrac{\ln 2}{10}\times 10 \\
\Rightarrow \dfrac{2}{T'-T}=2 \\
\therefore T'=\dfrac{3}{2}T $
So, the correct option is B.
Note: Newton’s law of cooling states that, the rate at which a body placed in an open air say exposed to atmosphere changes its temperature by loss of heat to the surrounding which is directly proportional to the difference between the object’s temperature and its surroundings, given that the difference is small. As per this law rate of cooling is dependent upon the temperature, suppose we want a hot cup of tea to cool down, a cup of hot coffee will cool more quickly if we put it in the refrigerator.
Complete step by step answer:
Newton's Law of Cooling, rate of heat transfer is given by
\[\dfrac{dQ}{dt}=kA(T-{{T}_{0}})\]
The surrounding temperature is constant at T. The body takes time 10 minutes to cool from 3T to T. Writing this mathematically, from Newton’s law of cooling,
$\ln \dfrac{3T-T}{2T-T}=kt \\
\Rightarrow \ln 2=10k \\
\therefore k=\dfrac{\ln 2}{10} \\$
Now for next 10, minutes
$\ln \dfrac{2T-T}{T'-T}=kt \\
\Rightarrow \ln \dfrac{2}{T'-T}=\dfrac{\ln 2}{10}\times 10 \\
\Rightarrow \dfrac{2}{T'-T}=2 \\
\therefore T'=\dfrac{3}{2}T $
So, the correct option is B.
Note: Newton’s law of cooling states that, the rate at which a body placed in an open air say exposed to atmosphere changes its temperature by loss of heat to the surrounding which is directly proportional to the difference between the object’s temperature and its surroundings, given that the difference is small. As per this law rate of cooling is dependent upon the temperature, suppose we want a hot cup of tea to cool down, a cup of hot coffee will cool more quickly if we put it in the refrigerator.
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