A body cools at the rate `0.5^(@)C//` minute when is `25^(@)C` above the surroundings. Calculate the rate of cooling when it is `15^(@)C` above the same surroundings.

Explanation

Given : `(d theta_(1))/(dt) = 0.5^(@)C`/min at
`theta_(1)-theta_(0)=25^(@)C.`
`and theta_(2)-theta_(0)=15^(@)C`
According to Newton's law of cooling,
Rate of cooling, `(d theta)/(dt)=k(theta-theta_(0))`
`therefore ((d theta_(1))/(dt))/((d theta_(2))/(dt))=(theta_(1)-theta_(0))/(theta_(2)-theta_(0))`
Then, `(0.5)/((d theta_(2))/(dt))=(25)/(15)`
`therefore (d theta_(2))/(dt)=(0.5xx15)/(25)=0.3^(@)C`/min.