A body cools from `80^(@)C` to `50^(@)C` in 5 min-utes Calculate the time it takes to cool from `60^(@)C` to `30^(@)C` The temperature of the surroundings is `20^(@)C` .

Options

A. 9 min
B. 7 min
C. 8 min
D. 10min

Answer

A

Explanation

Given , initial temperature `T_(1)=80^(@)C`
Final temperature `T_(2)=50^(@)C`
temperature of the surroundings `T_(0)=20^(@)C`
`t_(1)=5"min"`
According to Newton ' S law of cooling .
Rate of cooling `(dT)/(dt)=K[(T_(1)+T_(2))/(2)-T_(0)]`
`((80-50))/(5)=k[(80+50)/(2)-20]`
`(30)/(5)=k(65-20)rArr6=kxx45`
`"or "k=(6)/(45)=(2)/(15)`
In second condition ,
Initial temperature `T'_(1)=60^(@)C`
Final temperature `T'_(2)=30^(@)C`
`t'` =?
Now , `((60-30))/(t')=(2)/(15)((60+30)/(2)-20)`
`(30)/(t')=(2)/(15)(45-20)"or "t'=(30xx15)/(2xx25)=9 "min"`