A body originally at $80$ degree Celsius cools down to $60$ degree Celsius in $20$ minutes.The temperature of air being $40$ degree Celsius. What will be the temperature of the body after $40$ minutes from the original ?
2026-04-02 23:25:07.1775172307
Newtons law of cooling.
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According to the Newton's law of cooling, you have to solve the differential equation $$T'(t)=-r(T(t)-T_{\text{air}})$$ where $T(t)$ is the temperature of the body at at time $t$ and $r$ is some positive constant. The solution is $$T(t)=T_{\text{air}}+(T(0)-T_{\text{air}}))e^{-rt}.$$ Since $T_{\text{air}}=40^{\circ}$ and $T(0)=80^{\circ}$, it follows that $$T(t)=40+40e^{-rt}.$$ If $T(20)=60^{\circ}$, what is $T(40)$?