A body cools from 50^(@)C to 49.9^(@)C i...

A body cools from `50^(@)C` to `49.9^(@)C` in 5 s. the temperature of the surroundings is `30^(@)C`. How long will the body, take to cool from `40^(@)C` to `39.9^(@)C`?

2.5 seconds

5 seconds

7.5 seconds

10 seconds

Text Solution

Verified by Experts

The correct Answer is:

`((d theta)/(dt))_(1)=k(theta_(AV)-theta_(0))`
`((d theta)/(dt))_(2)=k(theta_(AV)-theta_(0))`

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