A) \[0{}^\circ C\]
B) \[40{}^\circ C\]
C) \[80{}^\circ C\]
D) Less than \[0{}^\circ C\]
Correct Answer: A
Solution :
[a] Heat taken by ice to melt at \[0{}^\circ C\] is \[{{Q}_{1}}=mL=540\times 80=43200\,cal\] Heat given by water to cool up to \[0{}^\circ C\] is \[{{Q}_{2}}=ms\Delta \theta =540\times 1\times (80-0)=43200\,cal\]Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is \[0{}^\circ C.\] Short trick: For these types of frequently asked questions you can remember the following formula \[{{\theta }_{mix}}=\frac{{{m}_{w}}{{\theta }_{w}}-\frac{{{m}_{i}}{{L}_{i}}}{{{c}_{w}}}}{{{m}_{i}}+{{m}_{w}}}\] (See theory for more details) If \[{{m}_{w}}={{m}_{i}}\] then \[{{\theta }_{mix}}=\frac{{{\theta }_{w}}-\frac{{{L}_{i}}}{{{c}_{w}}}}{2}=\frac{80+\frac{80}{1}}{0}=0{}^\circ C\]
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