A body cools from `60^(@)C` to `50^(@)C` in 10 minutes . If the room temperature is `25^(@)C` and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next 10 minutes will be

Options

A. `38.5^(@)C`
B. `40^(@)C`
C. `42.85^(@)C`
D. `45^(@)C`

Answer

C

Explanation

We use `(dT)/(dt) = -k(T-T_(0))`
`int_(60)^(50)(dT)/(T-T_(0))=-kint_(0)^(600) dt`
Here `T_(0) = 25^(@)C`
`In ((50-25)/(60-25)) =-k(600)`…(1)
In next 10minutes,Lettemperatureof bodyis `T^(@)C`
`int_(50)^(T)(dT)/(T -T_(0)) =-kint_(600)^(1200)dt`
`In ((T-25)/(50-25)) = -k(1200-600)`
=-600 K ...(2)
From (1) and (2), we get
`In((50-25)/(60-25))=In((T-25)/(50-25))`
`(25)/(35) =(T-25)/(25)`
`rArr T = 42.85^(@)C`