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` .

Explanation

According to Newton.s law of cooling we can write Rate of loss of heat `prop (T - T_0)`
`T_0` = Temperature of surroundings = `20^@C`
Let `T_1` = Initial temperature of body = `80^@C`
`T_2` = Final temperature of body `= 50^@C`
T = Mean Temperature =` (80 + 50)/(2) = 65^@C`
` t = 5 "min" = 5 xx 60 = 300 s`
`(dT)/(t) = k (T - T_0) `
` (mc (T_1 - T_2 ))/(t) = k (T - T_0)`
`(mc (80 - 50 ))/(300) = k (65 - 20)` ....(i)
Let t be the time taken by the body to cool from `60^@C` to `30^@C`
` therefore T = (60+30)/(2) = 45^@C`
`(mc (60 - 30 ))/(t ) = k (45 - 20)` ....(ii)
Dividing (i) by (ii) we get
`(30 xx t)/(300 xx 30 ) = 45/25 `
`t = (45 xx 300 xx 30)/(25 xx 30) = 540 s `