Question

A body cools according to Newton’s law from 100° C to 60° C in 20 minutes. The temperature of the surrounding being 20° C. How long will it take to cool down to 30° C?

Answer 1

Let θ°C be the temperature of the body at time t. The temperature of the surrounding is given to be 20° C.

According to Newton’s law of cooling

`("d"theta)/"dt" prop theta - 20`

∴ `("d"theta)/"dt" = - "k" (theta - 20)`, where k > 0

∴ `("d"theta)/(theta - 20) = - "k dt"`

On integrating, we get

`int 1/(theta - 20) "d"theta = - "k" int "dt" + "c"`

∴ log (θ - 20) = - kt + c

Initially, i.e. when t = 0, θ = 100

∴  log (100 - 20) = - k × 0 + c      ∴ c = log 80

 ∴ log (θ - 20) = - kt + log 80

∴ log (θ - 20) - log 80 = - kt

∴ `log ((theta - 20)/80) = - "kt"`    ....(1)

Now, when t = 20, θ = 60

∴ `log ((60 - 20)/80) = - "k" xx 20`

∴ `log (40/80) = - 20 "k"`

∴ k = `- 1/20 log (1/2)`

∴ (1) becomes, `log ((theta - 20)/80) = "t"/20 log (1/2)`

When θ = 30, then

`log ((30 - 20)/80) = "t"/20 log (1/2)`

∴ `log (1/8) = log (1/2)^("t"/20)`

∴ `(1/2)^("t"/20) = 1/8 = (1/2)^3`

∴ `"t"/20 = 3`     ∴ t = 60

∴ the body will cool down to 30° C in 60 minutes, i.e. in 1 hour.