Q. 37

# Two bodies of mas

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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