Answer :

Given length of metal rod = L


specific heat capacity of water = s


Mass of water =m


Rate of transfer of heat –


=


= is change in temperature.


A= Area of cross section of the tube


K = thermal conductivity of the tube


L= length


In time ∆t, the heat transfer from the rod will be given by


∆Q= (1)


Now, heat loss by water at temperature T1 is equal to the heat gain by water at temperature T2


So, heat loss by water at temperature T1 in time ∆t is -


∆Q=ms(T1-T1’) (2)


Where


m = mass of water


S = specific heat of water


From (1) and (2)


ms(T1-T1’) =


T1’=


This is the fall in temperature of water at temperature T1.


Similarly, rise in temperature of water at temperature T2


T2’=


Finally, change in temperature is given by-


(T1’-T2’)=


{(T1’-T2’)- (T1-T2)} = -


= -


Where is the rate of change of temperature difference.


Taking integral on both sides-


= dt


t =


Hence the time taken for the difference between the temperatures in the vessels to become half of the original value is


t =


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