A) \[{{31.2}^{o}}C\]
B) \[{{32.8}^{o}}C\]
C) \[{{36.7}^{o}}C\]
D) \[{{38.2}^{o}}C\]
Correct Answer: D
Solution :
Let neat given by water to cool upto\[{{0}^{o}}C=mc\Delta \theta \] where \[m\] is mass, \[c\] is specific heat, \[\Delta \theta \] is temperature difference. Heat taken by ice to melt\[=mL\]where \[L\] is latent heat Also if \[\theta \] is the temperature of the mixture then. Heat taken = Heat given \[mc\Delta \theta =mL+mc\Delta \theta \] \[100\times 1\times (50-\theta )=10\times 80+10\times 1\times (\theta -0)\] \[\Rightarrow \] \[500-100=80+0\] \[\Rightarrow \] \[11\theta =420\] \[\Rightarrow \] \[\theta =\frac{420}{11}={{38.2}^{o}}C\]
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