InterviewSolution
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InterviewSolution
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A slab of stone of area of `0.36 m^(2)` and thickness `0.1 m` is exposed on the lower surface to steam at `100^(@)C`. A block of ice at `0^(@)C` rests on the upper surface of the slab. In one hour `4.8 kg` of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice `= 3.63 xx 10^(5) J kg^(-1)`)A. `1.29J//m//s//^(@)C`B. `2.05J//m//s//^(@)C`C. `1.02J//m//s//^(@)C`D. `1.24J//m//s//^(@)C` |
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Answer» Correct Answer - D `(dQ)/(dt)=(KA)/(L)(T_(1)-T_(2))` `Q=(KA)/(L)(T_(1)-T_(2))t` `Q=mL_(f)` `(KA)/(L)(T_(1)-T_(2)t=mL_(f)` `K=(mL_(f)(L))/(A(T_(1)-T_(2))t)` `K=(4.8xx3.36xx10^(5)xx0.1)/(0.36xx100xx3600)J//m//s//^(@)C` `=(4.8xx3.36)/(0.36xx36)` `1.24J//m//s//^(@)C` |
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