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# A body cools from ${70}^{\circ }\text{c}$  to  ${50}^{\circ }\text{c}$ in 6 minutes. If temperature of surroundings is  ${30}^{\circ }\text{c}$.  What is the temperature after 12 minutes?

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a
20∘c
b
30∘c
c
34∘c
d
40∘c
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detailed solution

Correct option is C

initial temperature of the body is θ1=70oc  after 6 minutestemperature of body is θ2=50oc; surrounding temperature is     θ0=30oc; final temperature of body is    θ =? time t1=6 minutestime t2=12 minutesfrom newton law of cooling  dθdt=-kθ1+θ22-θ0---(1)  substitute given values in eqn(1)      θ1-θ2time=-kθ1+θ22-θ070-506=-k70+502-30206=-k60-30k=-19---(2)  substitute second part of question, given values and eqn(2) in eqn(1)50-θ12=--1950+θ2-30 ⇒θ=340C

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