10 g of ice at $- 20^{o} C$ is dropped into a calorimeter containing 10 g of water at $10^{o} C$ ; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain:

20 g of water

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20 g of ice

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10 g ice and 10 g water

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5 g ice and 15 g water

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Solution

The correct option is C 10 g ice and 10 g water
According to principle of calorimetry,
$H e a t g a i n e d b y i c e = (10 \times 80) + 10 \times 0.5 (0 - (- 20))$
$800 + 100 = 900 c a l$
$H e a t l o s t b y w a t e r = 10 \times 1 \times (10 - 0) + 80 \times 10$
$= 900 c a l$
$∴ H e a t l o s t = H e a t g a i n e d$
At equilibrium, the calorimeter will contain 10 g of ice and 10 g of water.

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10gm of ice at $- 20^{\circ}$ is dropped into a calorimeter containing 10gm of water at $10^{\circ} C$ , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain.