A metal sphere cools at the rate of `1.6^@ C`/min when its temperature is `70^@ C`. At what rate will it cool when its temperature is `40^@ `C. The temperature of surroundings is `30^@ `C.

Explanation

Given ` T_1 = 70^@ C`
`T_2 = 40^@ C`
` T_0 = 30^@ C`
` ((dT)/(dt))_1 = 1.6^@ C // min`
According to Newton’s law of cooling, if C is the constant of proportionality
`((dT)/(dt))_1 = C (T_1 -T_0)`
or `1.6 = C ( 70-30)`
` therefore C = ( 1.6 )/(40 ) = 0.04 // min`
Also` ((dT)/(dt))_2 = C (T_2 -T_0)`
`= 0.04( 40-30)= 0.4^@ C // min`
Thus the rate of cooling drops by a factor of four when the difference in temperature of the metal sphere and its surroundings drops by a factor of four.