A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^(@)C`. Initial temeprature of cooling is `40^@C` and assuming newtons law of cooling is valid. The body cools to `38^@C` in 10 min
In further 10 min it will cools from `38^@C` to _____

Options

A. `36^(@)C`
B. `36.4^(@)C`
C. `37^(@)C`
D. `37.5^(@)C`

Answer

B

Explanation

We have `theta-theta_(s)=(theta_(0)-theta_(s))e^(-kt)" "….(1)`
where `theta_(0)=` initial temperature of body `=40^(@)C`
`theta=` temperature of body after time t
`theta_(s)=` temperature of surrounding
Since body cools from `40^(@)C` to `38^(@)C` in 10 min, we have
`38^(@)-30^(@)=(40^(@) -30^(@))e^(-10k)" "...(2)`
Let after 10 min, the body temp. be `theta`.
`theta-30^(@)=(38^(@)-30^(@))e^(-10k)" " ...(3)`
Dividing equ. (2) by equ. (3) gives,
`(8^(@))/(theta-30^(@))=(10^(@))/(8^(@)) rArr theta-30^(@)=6.4^(@) rArr theta=36.4^(@)`