A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^(@)C`. Initial temperature of cooling is valid. The body of mass 1 kg cools to `38^@C` in 10 min
When the body temperature has reached `38^@C`, it is heated again so that it reaches `40^@C` in 10 min. The heat required from a heater by the body is

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)`
Where `theta_0=`initial temperature of body `=40^@C`
`theta=`temperature of body after time t.
Since body cools from 40 to 38 in 10 min, we have
`38-30=(40-30)e^(-10k)` .(i)
Let after 10 min, the body temperature be `theta`
`theta-30^@=(38-30)e^(-10k)` .(ii)
`(Eq(i))/(Eq(ii))`gives `(8)/(theta-30)=(10)/(8)`,`theta-30=6.4`
`theta=36.4^@C`