A body cools at the rate of 3^(@)C//min ...

A body cools at the rate of `3^(@)C//`min when its temperature is `50^(@)C`. If the temperature of surroundings is `25^(@)C` then the rate of cooling of the body at `40^(@)C` is

`2^(@)C`/min

`2.4^(@)C`/min

`2.8^(@)C`/min

`1.8^(@)C`/min

Text Solution

Verified by Experts

The correct Answer is:

`R_(1)=k(theta_(1)-theta_(0))`
`R_(2)=k(theta_(2)-theta_(0))`
`(R_(2))/(R_(1))=(40-25)/(50-25)=(15)/(25)=(3)/(5)`
`R_(2)=(3)/(5)xx3=1.8^(@)C//`min.

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