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60 g of ice at 00C is mixed with 60 g of steam at 1000C. At thermal eqilibrium, the mixture contains
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a
80 g of water and 40 g of steam at 1000C
b
120 g of water 900C
c
120 g of water at 1000C
d
40 g of steam and 80 g of water at 00C
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Correct option is A
mLS = Q1Q1 = 60 x 540 = 32400 → steam to water at 1000CmLi + msΔt = Q2Q2 = 60 x 80 + 60 x 1 x 100 = 10800 → ice to water at 1000CSince Q1 > Q2, a part of the steam is condensed.If x gm of steam is converted into water at 1000C so the final mixture contains (60 - 20)gm of steam is 40gm of steam and (60 + 20) gm of water i.e. 80gm of water.Similar Questions
10 g of ice at -20°C is dropped into a calorimeter containing 10 g of water at l 0°C; the specific heat _of water is twice that of ice . When equilibrium is reached,· the calorimeter will contain:
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