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One end of the rod of length l, thermal ...

One end of the rod of length l, thermal conductivity K and area of cross-section A is maintained at a constant temperature of `100^@C`. At the other end large quantity of ice is kept at `0^@C` . Due to temperature difference, heat flows from left end to right end of the rod. Due to this heat ice will start melting. Neglecting the radiation losses find the expression of rate of melting of ice.
.

Text Solution

Verified by Experts

The correct Answer is:
A, D
Putting,
`((dQ)/dt)_("conduction")=((dQ)/(dt))_("calorimetry")`
`:. (TD)/R = L* (dm)/(dt)`
`rArr (dm)/(dt) = (TD)/(RL)`
So, this is the desired expression of `(dm)/(dt)`.
In the above expression,
`TD = 100^@C, R = l/(KA)`
and L = latent heat of fusion.
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Knowledge Check

  • One end of conducting rod is maintained at temperature 50^(@)C and at the other end ice is melting at 0^(@)C . The rate of melting of ice is doubled if:

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    the temperature is made `200^(@)C` and the area of cross-section of the rod is doubled
    B
    the temperature is made `100^(@)C` and the length of the rod is made four time
    C
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    D
    the temperature is made `100^(@)C` and area of cross-section of rod and length both are doubled.

  • The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is 4 cal//s . Then, in the second rod rate of heat flow in cal//s will be

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  • One end of a metal rod of length 1.0 m and area of cross section 100 cm^(2) is maintained at . 100^(@)C . If the other end of the rod is maintained at 0^(@)C , the quantity of heat transmitted through the rod per minute is (Coefficient of thermal conductivity of material of rod = 100 W//m-K )

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