A) 0°C
B) 40°C
C) 80°C
D) Less than 0°C
Correct Answer: A
Solution :
Heat taken by ice to melt at 0°C is \[{{Q}_{1}}=mL=540\times 80=43200\,cal\] Heat given by water to cool upto 0°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°C. Short trick : For these type of frequently asked questions you can remember the following formula \[{{\theta }_{\text{mix}}}=\frac{{{m}_{W}}{{\theta }_{W}}-\frac{{{m}_{i}}{{L}_{i}}}{{{c}_{W}}}}{{{m}_{i}}+{{m}_{W}}}\] (See theory for more details) If \[{{m}_{W}}={{m}_{i}}\] then \[{{\theta }_{mix}}=\frac{{{\theta }_{W}}-\frac{{{L}_{i}}}{{{c}_{W}}}}{2}=\frac{80-\frac{80}{1}}{2}=0{}^\circ C\]
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