Water placed in a container cools from `80^@C` to `70^@C` in 5 minutes. How long will it take to cool from `60^@C` to `40^@C` if temperature of surroundings is `12^@C`

Explanation

Case (i) `T = (80 + 70)/(2) = 75^@C , dT = 80 - 70 = 10^@C`
dt = 5 min
`T_0 = 12^@C`
` (dT)/(dt) = K (T - T_0) `
` 10/5 = K (75 - 12) ` …..(i)
case (ii) `T = (60+40)/(2) = 50^@C , dT = 60 - 40 = 10^@C`
`T_0 = 12^@C`
` T_0 = 12^@C`
`(dT)/(dt) = K (T - T_0) `
`(50)/(dt) = K (50 - 12) ` ....(ii)
Dividing (i) by (ii)
`(10 xx dt)/(5 xx 50) = (K (75 - 12))/(K (50 - 12) ) `
dt = 41.41 min