Answer:
Let \[{{m}_{1}}\] and \[{{m}_{2}}\] be the masses of ice melted in same time \[t\text{ }(=\text{1 min})\]in vessels A and B respectively. Then the amounts of heat flowed into the two vessels will be \[{{Q}_{1}}=\frac{{{K}_{1}}A({{T}_{1}}-{{T}_{2}})t}{x}={{m}_{1}}L\] ?(i) \[{{Q}_{2}}=\frac{{{K}_{2}}A({{T}_{1}}-{{T}_{2}})t}{x}={{m}_{2}}L\] ?(ii) where L is latent heat of ice. Dividing (i) by (ii), we get \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{100g}{150g}=\frac{2}{3}=\mathbf{2:3}\]
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