∴ Rate of loss of heat is given by dQ dt ms dT dt ( . ) From Eqs. ( . ) and ( .
) we have m s dT dt k T dT k ms dt K dt ( . ) where K = k/m s On integrating, log e ( T – T ) = – K t + c ( . ) or T = T + C ′ e – Kt ; where C ′ = e c ( . ) Equation ( .
) enables you to calculate the time of cooling of a body through a particular range of temperature. For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table. Fig.
. Verification of Newton’s Law of cooling. Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. .
(a). The set-up consists of a double- walled vessel (V) containing water between the two walls. A copper calorimeter (C) containing hot water is placed inside the double-walled vessel. Two thermometers through the corks are used to note the temperatures T of water in calorimeter and T of hot water in between the double walls, respectively.
Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between log e ( T – T ) [or l n( T – T )] and time ( t ). The nature of the graph is observed to be a straight line having a negative slope as shown in Fig. .
(b). This is in support of Eq. . .
Example . A pan filled with hot food cools from ° C to ° C in minutes when the room temperature is at ° C. How long will it take to cool from ° C