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. `42.85^(@)C`
B. `45^(@)C`
C. `40.46^(@)C`
D. `44.23^(@)C`

Explanation

(a) According to Newton's law of cooling,
`(d theta)/(dt)=K(theta-theta_(0))`
where `theta=` temperature of body
`theta_(0)=` temperature of surroundings
`int_(theta_(1))^(theta_(2))(d theta)/(theta-theta_(0))=-Kint_(0)^(1)dt`
`log ((theta_(2)-theta_(0))/(theta_(1)-theta_(0)))=-Kt`
`log((60-25)/(50-25))=-Kxx10xx60`
Let `alpha=` temperature at end of 10 minutes
`log((50-25)/(alpha-25)=Kxx10xx60`
`alpha=42.85^(@)C`