A) \[~10{{\,}^{o}}C\]
B) \[~27{{\,}^{o}}C\]
C) \[~14{{\,}^{o}}C\]
D) none of these
Correct Answer: A
Solution :
Key Idea: Use Calorimetrys principle. When ice is mixed with water, then heat given by water, \[{{Q}_{1}}={{m}_{1}}{{s}_{1}}({{t}_{1}}-t)\] where t is equilibrium temperature. Heat taken by ice to melt, \[{{Q}_{2}}={{m}_{2}}{{s}_{2}}(t-{{t}_{2}})+{{m}_{2}}L\] From principle of Calorimetry. Heat given by water = Heat taken by ice i.e., \[{{m}_{1}}{{s}_{1}}({{t}_{1}}-t)={{m}_{2}}{{s}_{2}}(t-{{t}_{2}})+{{m}_{2}}L\] Given, \[{{m}_{1}}={{m}_{2}}=100\,g,\,{{t}_{1}}=100{{\,}^{o}}C,\] \[{{t}_{2}}=0{{\,}^{o}}C,{{s}_{1}}=1\,cal/{{g}^{o}}C,{{s}_{2}}=1\,cal/{{g}^{o}}C,\] \[L=80\,cal/g\] Substituting the values in Eq. (i), we get \[100\times (100-t)=100\times 1\times (t-0)+100\times 80\] or \[10000-100\,t=100\,t+8000\] or \[200\,t=2000\] or \[t=10{{\,}^{o}}C\]
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