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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
$10\, gm$ of ice at $-20°C$ is dropped into a calorimeter containing $10\, gm$ of water at $10°C;$ the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain
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
Solution
(c) Heat given by water ${Q_1} = 10 \times 10 = 100\,cal.$
Heat taken by ice to melt
$Q2 = 10 \times 0.5 \times [0 -(-20)] + 10 \times 80 = 900 cal$
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 $0°C.$
Standard 11
Physics
