A) 20 gm of water
B) 20 gm of ice
C) 10 gm ice and 10 gm water
D) 5 gm ice and 15 gm water
Correct Answer: C
Solution :
Heat given by water \[{{Q}_{1}}=10\times 10=100\,cal.\] Heat taken by ice to melt Q2 = 10 ´ 0.5 ´ [0 ? (? 20)] + 10 ´ 80 = 900 cal As \[{{Q}_{1}}<{{Q}_{2}},\] so ice will not completely melt and final temperature = 0°C. As heat given by water in cooling up to 0°C is only just sufficient to increase the temperature of ice from ? 20°C to 0°C, hence mixture in equilibrium will consist of 10 gm ice and 10 gm water at 0°C.
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