$50 g$ of ice at $- 5^{o} C$ is added to $200 g$ of water at $40^{o} C$ in a calorimeter. The water equivalent of calorimeter is $50 g$ of water. The temperature of mixture is

{9.58}^{o} C

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{12.58}^{o} C

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{19.58}^{o} C

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{29.58}^{o} C

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Solution

The correct option is B ${19.58}^{o} C$
Heat absorbed by water and calorimeter to reach $0^{o} C$ is $200 \times 4.2 \times 40 + 50 \times 4.2 \times 40 = 42000 J$
Heat absorbed by ice when it melts is $50 \times 2.1 \times 5 + 50 \times 336 = 17325$
so, $42000 - 17325 = (50 + 200 + 50) \times 4.2 \times (T - 0)$
$T = {19.58}^{o} C$

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5 g m

of water at

30^{\circ} C

and

5 g m

of ice at

- 20^{\circ} C

are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice

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= 80 c a l / g m

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50 g of ice at −5oC is added to 200 g of water at 40oC in a calorimeter. The water equivalent of calorimeter is 50 g of water. The temperature of mixture is

The correct option is B 19.58oCHeat absorbed by water and calorimeter to reach 0oC is 200×4.2×40+50×4.2×40=42000 JHeat absorbed by ice when it melts is 50×2.1×5+50×336=17325so, 42000−17325=(50+200+50)×4.2×(T−0)T=19.58oC

$50 g$ of ice at $- 5^{o} C$ is added to $200 g$ of water at $40^{o} C$ in a calorimeter. The water equivalent of calorimeter is $50 g$ of water. The temperature of mixture is