A body initially at `80^(@)C` cools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What will be its temperature in 15 minutes and what is the temperature of its surroundings?

Options

A. `42.7^(@)C`
B. `35^(@)C`
C. `47^(@)C`
D. `40^(@)C`

Answer

A

Explanation

From Newton.s law of cooling:-
`(theta_(1)-theta_(2))/(t)=k((theta_(1)+theta_(2))/(2)-theta_(0))` where `theta_(1)` is higher temperature, `theta_(2)` is lower temperature.
`(80-64)/(5)=k(72-theta_(0))" "…(i)`
Where `theta_(0)` is temperature of surroundings
`(64-52)/(10)=k(58-theta_(0)) " " ...(ii)`
Dividing (i) and (ii) we get `theta_(0)`
`(52-theta)/(15)=k((52+theta)/(2)-theta_(0))" " ...(iii)`
Thus `theta` is found.