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Question: A body at temperature \({{40}^{0}}C\) is kept in a surrounding of constant temperature \({{20}^{0}}C...

A body at temperature 400C{{40}^{0}}C is kept in a surrounding of constant temperature 200C{{20}^{0}}C ,it is observed that its temperature fasts to 350C{{35}^{0}}C find how much more time it will take for the body to attain a temperature of 300C{{30}^{0}}C
A) 14 minutes
B) 28 minutes
C) 56 minutes
D) 125 minutes

Explanation

Solution

This problem solved by newton’s law of cooling ,newton’s law of cooling can be derived by Stefan –Boltzmann law the rate of cooling is defined as decreasing temperature per unit area and newton’s law of cooling holds good irrespective of mode of heat transfer (conduction, convection or radiation)

Complete step by step answer:
Newton’s law of cooling states that the rate of heat loss of a body is proportional to the difference in temperature between body and its surrounding i.e,
dQdt=k1(TTs)\dfrac{dQ}{dt}=-{{k}_{1}}(T-{{T}_{s}})
Where TT is the temperature of body
Ts{{T}_{s}} is the temperature of surrounding
kk is the heat transfer coefficient
The rate of cooling is defined as decrease in temperature per unit time (dTdt)(\dfrac{dT}{dt}) substitute dQ=msdTdQ=msdT in the above equation to get useful forms of newton’s law of cooling
dTdt=k1(TT0)\dfrac{dT}{dt}=-{{k}_{1}}(T-{{T}_{0}})
The differential equation is solved by using initial conditions. If T0{{T}_{0}} is the initial temperature of
the body then its temperature at time t is given by
Δθ1=Δθ1ekt\Delta {{\theta }_{1}}=\Delta {{\theta }_{1}}{{e}^{-kt}} (1)\cdots \cdots (1)
For the interval in which temperature falls form 400C{{40}^{0}}C to 350C{{35}^{0}}C
Substituting in equation (1)(1)
(300200)=(400200)ekt({{30}^{0}}-{{20}^{0}})=({{40}^{0}}-{{20}^{0}}){{e}^{-kt}}
e10k=34{{e}^{-10k}}=\dfrac{3}{4}
k=ln4310k=\dfrac{\ln \dfrac{4}{3}}{10}
For the next internal
equation(1)(1) becomes
(300200)=(350200)ekt({{30}^{0}}-{{20}^{0}})=({{35}^{0}}-{{20}^{0}}){{e}^{-kt}}
ekt=23 kt=ln32 \begin{aligned} & {{e}^{-kt}}=\dfrac{2}{3} \\\ & kt=\ln \dfrac{3}{2} \\\ \end{aligned} (3)\cdots \cdots (3)
Equating equation(2)(2) and (3)(3)
(ln43)t10=ln32\dfrac{(\ln \dfrac{4}{3})t}{10}=\ln \dfrac{3}{2}
t=10(ln32)(ln43)t=10\dfrac{(\ln \dfrac{3}{2})}{(\ln \dfrac{4}{3})}
t=14.09mint=14.09\min
The time it will take for the body to attain a temperature of 300C{{30}^{0}}C is 14.09min

So, the correct answer is “Option A”.

Note: Students if the difference in temperature between the body and its surrounding is not large then newton’s law of cooling holds good. The law is frequently qualified to include the condition that the temperature difference is small and the nature of heat transfer mechanism remains the same; the amount of thermal energy in the body is calculated by assuming heat capacity.