A) 86g
B) 103g
C) 77g
D) 63g
Correct Answer: A
Solution :
Given, total mass of water =M= 100 g \[{{L}_{1}}=2.1\times {{10}^{5}}\text{J}\,\text{k}{{\text{g}}^{-1}}\] \[{{L}_{2}}=3.36\times {{10}^{5}}\text{J}\,\text{k}{{\text{g}}^{-1}}\] Suppose mass of ice formed = m the mass of water evaporated = M ? m amount of heat taken by water evaporate \[=(M-M){{L}_{1}}\] On freezing heat given by water\[=m{{L}_{2}}\] As, \[M{{L}_{2}}=(M-m){{L}_{1}}\] \[\Rightarrow \]\[m=\frac{M{{L}_{1}}}{{{L}_{1}}+{{L}_{2}}}\]\[\Rightarrow \]\[m=\frac{100(21\times {{10}^{5}})}{(21+3.6)\times {{10}^{5}}}=86g\]You need to login to perform this action.
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