Q. 36

# Figure shows two

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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