A body cools from a temperature `3 T` to `2 T` 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 next `10` minutes will be

Options

A. `(7)/(4)T`
B. `(3)/(2)T`
C. `(4)/(3)T`
D. T

Answer

B

Explanation

According to Newton.s law of cooling, `(dT)/(dt) = K (T-T_(s))`
For two cases, `(dT_(1))/(dt) = K (T_(1) - T_(s)) and (dT_(2))/(dt) = K (T_(2) - T_(s))`
Here, `T_(s) = T, T_(1) = (3T + 2T)/(2) = 2.5T`
and `(dT_(1))/(dt) = (3T-2T)/(10) = (T)/(10)`
`T_(2) = (2T-T.)/(2) and (dT_(2))/(dt) = (2T. - T.)/(10)`
So, `(T)/(10) = K (2.5T - T)` ...(i)
`(2T- T.)/(10) = K ((2T+ T.)/(2) - T)`...(ii)
Dividing eqn. (i) by eqn (ii) we get
`(T)/(2T-T.) = ((2.5T- T))/(((2T +T.)/(2))-T)`
`(2T+ T.)/(2) -T = (2T- T.) xx (3)/(2)`
`T. = 3(2T-T.) or, 4T. = 6T :. T.= (3)/(2)T`