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Standard X

Physics

Calorimeter

1 Kg of ice a...

Question

$1 K g$ of ice at $- 20^{\circ} C$ is mixed with $2 K g$ of water at $90^{\circ} C$ . Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume latent heat of ice $= 334.4 K J / K g$ , specific heat of water and ice are $4.18 k J / (k g . K)$ and $2.09 k J / (k g . K)$ , respectively.)

30^{\circ} C

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0^{\circ} C

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80^{\circ} C

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45^{\circ} C

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Solution

The correct option is A $30^{\circ} C$
Heat gained by ice =heat loss by water
$1 \times 2.09 \times 20 + 1 \times 334.4 + 1 \times 4.18 \times T = 2 \times 4.18 \times (90 - T)$
$T = 30^{o}$

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The correct option is A 30∘CHeat gained by ice =heat loss by water 1×2.09×20+1×334.4+1×4.18×T=2×4.18×(90−T)T=30o

The correct option is A $30^{\circ} C$
Heat gained by ice =heat loss by water
$1 \times 2.09 \times 20 + 1 \times 334.4 + 1 \times 4.18 \times T = 2 \times 4.18 \times (90 - T)$
$T = 30^{o}$