A mixture containing 22.4 g of ice (at exactly 0.00 ∘C) and 78.7 g of water (at 62.1 ∘C) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temperature of the mixture?

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A mixture containing 22.4 g of ice (at exactly 0.00 ∘C) and 78.7 g of water (at 62.1 ∘C) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temperature of the mixture?

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Answer 1 · 2018-07-14T09:15:39

Taking sp.heat of water to be $1 c a l g^{- 1}^{\circ} C^{- 1}$ $1calg^-1""^@C^-1$

If water reaches at $0^{\circ} C$ $0^@C$ final temperature then heat lost by water would be

$= 78.7 \times 1 \times 62.1 = 4887.27 c a l$ $=78.7xx1xx62.1=4887.27cal$

The heat required to melt ice at $0^{\circ} C$ $0^@C$ is $22.4 . \times 80 = 1792 c a l$ $22.4.xx80=1792cal$ which is much less than the heat lost.

So final temperature will be higher than $0^{\circ} C$ $0^@C$ . Let it be $t^{\circ} C$ $t^@C$ . So heat gained by ice cold water to reach at this temperature will be
$= 22.4 \times 1 \times t = 22.4 t$ $=22.4xx1xxt=22.4t$ Cal.

Again the heat lost by water at ${62.1}^{\circ} C$ $62.1^@C$ to reach the final temperature will be

$78.7 \times 1 \times (62.1 - t)$ $78.7xx1xx(62.1-t)$ Cal.

By calorimetric principle

$22.4 t + 1792 = 78.7 \times 1 \times (62.1 - t)$ $22.4t+1792=78.7xx1xx(62.1-t)$

$\Rightarrow 22.4 t + 78.7 t = 4887.27 - 1792$ $=>22.4t+78.7t=4887.27-1792$

$\Rightarrow 101.1 t = 4887.27 - 1792$ $=>101.1t=4887.27-1792$

$\Rightarrow t \approx {30.6}^{\circ} C$ $=>t~~30.6^@C$

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A mixture containing 22.4 g of ice (at exactly 0.00 ∘C) and 78.7 g of water (at 62.1 ∘C) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temperature of the mixture?

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