One end of a 0.25 m long metal bar is in steam and the other is in contact with ice . If 12 g of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar `=5xx10^(-4) m^(2)` and latent heat of ice is `80 cal g^(-1)`

Options

A. `20 cal s^(-1) m^(-1).^(@)C^(-1)`
B. `10 cal s^(-1) m^(-1) .^(@)C^(-1)`
C. `40 cal s^(-1) m^(-1) .^(@)C^(-1)`
D. `80 cal s^(-1) m^(-1) .^(@)C^(-1)`

Answer

D

Explanation

Here, `x = 0.25 m, T_(T) - T_(2) = 100 - 0 = 100 .^(2)C`
`t = 1` min `= 60 s, A = 5 xx 10^(-4) m^(2), L = 80 cal g^(-1)`
`Q = mL = 12 xx 80 = 960 cal`
But, `Q =(KA(T_(1) - T_(2))t)/(x)`
or `960 = (K xx 5 xx 10^(-4) xx 100 xx 60)/(0.25)`
`:. K = (960 xx 0.25)/(5 xx 10^(-2) xx 60) = 80 cal s^(-1) m^(-1) .^(@)C^(-1)`