Answer :

Given-


Masses of body = m1 and m2


Specific heat capacities = s1 and s2


Rod of length= ℓ,


Cross-sectional area = A


Thermal conductivity = K


Rate of transfer of heat from the rod is given by –


=(1)


Where,= temperature of first and second body.


A= Area of cross section of the


K = thermal conductivity of the


L= length


Heat transfer from the rod in time ∆t –


∆Q= (2)


Heat loss by the body at temperature T2 is equal to the heat gain by the body at temperature T1.


Heat loss by the body at temperature T2 in time ∆t is –


∆Q=m2s2 (T2′-T2) (3)


From (1) and (2)


m2s2 (T2′-T2) =


T2’=T2- ∆t


This is the fall in the temperature of the body at temperature T2.


Similarly, rise in temperature of water at temperature T1 is –


T1’=T1+ t


Change in the temperature


(T2′-T1′)


= T2- ∆t - T1- t


{(T2′-T1′)-(T2-T1)}=- ∆t - t


= -


Where is the rate of change of temperature difference


∆T= )


Integrating both the sides -


=


ln |T2-T1| = -


Taking the anti-log


(T2-T1)=e-λt


Where


λ=


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