A hot body is kept in cooler surroundings. Its rate of cooling is ` 3^(@)C`/min when its temperature is ` 60^(@)C and 1.5^(@)C`/ min when its temperature is ` 45^(@)C`. Determine the temperature of the surroundings and the rate of cooling when the temperature of the body is ` 40^(@)C`

Explanation

When ` theta_(2) = 45^(@) C , theta_(3) = 40^(@)C`
By Newton's law of cooling `(dtheta)/(dt) = k (theta- theta_(0))`
` 3 = k (60 - theta_(0)) and 1.5 = k (45- theta_(0))`
`3/1.5 = ((60 - theta_(0))/((45-theta_(0)))`
` 2 = ( 60 - theta_(0))/( 45- theta_(0))`
` 60 - theta _(0) = 90 - 2theta_(0)`
`theta_(0) = 30^(@)C`
THe temperature of the surroundings = ` 30^(@)C`
Now, ` ((d theta//dt)_(3))/((d theta//dt)_(2)) = (theta_(3) - theta_(0))/(theta_(2) - theta_(0))`
`((dtheta//dt)_(3))/(1.5) = (40 -30)/(45-30) `
` ((dtheta)/(dt))_(3) = 1.5 xx 10/15 = 1^(@)C`/min
The rate of cooling when the temperature of the body is ` 40^(@) C = 1^(@) C//min`