A block of ice weighing 20 gm, at - $10^{o} C$ is mixed with 100 gm water at
$10^{o} C$ , in an insulated flask. What is the amount of water, in the flask, when equilibrium is reached?

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Solution

Heat required to raise the temp. of ice to $0^{o} C = 20 \times 0.5 \times 10 = 100 c a l$
Heat supplied by water coming to $0^{o} C = 100 \times 1 \times 10 = 1000 c a l$
Remaining heat to melt ice $= 900 c a l$
Amount of ice that will melt $= 900 / 80 = 11.25 g m$
Total water amount at end $= 111.25 g m$

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A block of ice weighing 20 gm, at -10oC is mixed with 100 gm water at 10oC, in an insulated flask. What is the amount of water, in the flask, when equilibrium is reached?

Heat required to raise the temp. of ice to 0oC=20×0.5×10=100calHeat supplied by water coming to 0oC=100×1×10=1000calRemaining heat to melt ice =900calAmount of ice that will melt=900/80=11.25gmTotal water amount at end=111.25gm

A block of ice weighing 20 gm, at - $10^{o} C$ is mixed with 100 gm water at
$10^{o} C$ , in an insulated flask. What is the amount of water, in the flask, when equilibrium is reached?