A) \[{{10}^{o}}C\]
B) \[{{27}^{o}}C\]
C) \[{{14}^{o}}C\]
D) none of these
Correct Answer: A
Solution :
Key Idea: Use Calorimetry's 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 \[100\times (100-t)=100\times 1(t-0)+100\times 80\] or \[10000-100t=10t+8000\] or \[200t=2000\] or \[t={{10}^{o}}C\]
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