Answer :

Given-

Masses of body = m_{1} and m_{2}

Specific heat capacities = s_{1} and s_{2}

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 *T*_{2} is equal to the heat gain by the body at temperature *T*_{1}.

Heat loss by the body at temperature *T*_{2} in time ∆t is –

∆Q=m_{2}s_{2} (T_{2}′-T_{2}) (3)

From (1) and (2)

m_{2}s_{2} (T_{2}′-T_{2}) =

⇒ T_{2}’=T_{2}- ∆t

This is the fall in the temperature of the body at temperature *T*_{2}.

Similarly, rise in temperature of water at temperature *T*_{1} is –

T_{1}’=T_{1}+ t

Change in the temperature

(T_{2}′-T_{1}′)

= T_{2}- ∆t - T_{1}- t

⇒ {(T_{2}′-T_{1}′)-(T_{2}-T_{1})}=- ∆t - t

⇒ = -

Where is the rate of change of temperature difference

⇒∆T= )

Integrating both the sides -

=

⇒ ln |T_{2}-T_{1}| = -

Taking the anti-log

⇒ (T_{2}-T_{1})=e^{-λt}

Where

λ=

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