A body takes 10 min to cool douwn from `62^@C` to `50^@C`. If the temperature of surrounding is `26^@C` then in the next 10 minutes temperature of the body will be

Options

A. `38^(@)C`
B. `40^(@)C`
C. `42^(@)C`
D. `44^(@)C`

Answer

C

Explanation

From Newton's colling law `(theta_(1)-theta_(2))/t=K((theta_(1)+theta_(2))/2-26)`
Case I : `12/10 =Kxx30 rArr K=12/(10xx30) =1/25`
Case II: `(50 - theta_(2))/10 K((50 +theta_(2))/2-26) rArr (50 -theta_(2))/10=1/25((theta_(2)-2)/2)`
`250-5 theta_(2)=theta_(2)-2 rArr 6 theta_(2)=252`
`theta_(2)=252/6 =42^(@)C`