One end of a copper rod of length $1.0\;m$ and area of cross-section ${10^{ - 3}}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92\;cal/m{\rm{ - }}s{{\rm{ - }}^o}C$ and the latent heat of ice is $8 \times {10^4}cal/kg$, then the amount of ice which will melt in one minute is
Medium
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(c) Heat transferred in one minute is utilised in melting the ice
so, $\frac{{KA({\theta _1} - {\theta _2})t}}{l} = m \times L$
$ \Rightarrow m = \frac{{{{10}^{ - 3}} \times 92 \times (100 - 0) \times 60}}{{1 \times 8 \times {{10}^4}}} = 6.9 \times {10^{ - 3}}kg$
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