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A body cools at the rate of 2.8^(@)C/min...

A body cools at the rate of `2.8^(@)C`/min, when its temperature is `70^(@)C`. What will be the rate of cooling ( in `""^(@)C`/min) of the body at `40^(@)C` if the temperature of its surroundings is half the temperature of the body.

Text Solution

Verified by Experts

The correct Answer is:
`0.4`
`(d theta)/(dt) = K( theta-theta_(0))`
`:.2.8=K xx (70-35)=35K`
`:.(d theta)/(dt) =K xx (40-35) =5K`
`:. ((d theta)/(dt))/(2.8)=(5)/(35)`
`:. (d theta)/(dt) = 0.4^(@)C`/min
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