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Conduction

A lump of ice...

Question

A lump of ice of $0.1 k g$ at $- 10^{\circ} C$ is put in $0.15 k g$ of water at $20^{\circ} C$ . If $\frac{x}{100} k g$ of ice is found in the mixture when it has reached thermal equilibrium. The value of $x$ is _______.

[Given : Specific heat of ice $= 0.5 k c a l k g^{- 1} K^{- 1}$ and its latent heat of melting $= 80 k c a l / k g$ .]

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Solution

Heat released by $0.15 k g$ of water in being cooled to $0^{\circ} C = 0.15 \times 1 \times 20 = 3 k c a l$
Heat absorbed by ice from $- 10^{\circ} C$ to $0^{\circ} C = 0.1 \times 0.5 \times 10 = 0.5 k c a l$
The balanced heat is available for melting ice. Let $m k g$ of ice melt.
Then $m \times 80 = 2.5$ or $m = 0.03 k g$

Thus the final temperature is $0^{\circ} C$ with $0.07 k g$ of ice and $0.18 k g$ of water.

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0.10 k g

- 10^{\circ} C

0.15 k g

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= 1 k c a l k g^{- 1}

= 0.50 k c a l k g^{- 1}

= 80 k c a l k g^{- 1}

- 10^{\circ}

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(C_{i c e} = 0.5 k c a l k g^{- 1} k^{- 1}, L_{i c e} = 80 k c a l k g^{- 1})

0.10 k g

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= 1 k c a l k g^{- 1}

= 0.50 k c a l k g^{- 1}

= 80 k c a l k g^{- 1}

- 10^{\circ}

20^{\circ}

(C_{i c e} = 0.5 k c a l k g^{- 1} k^{- 1}, L_{i c e} = 80 k c a l k g^{- 1})

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1 Kcal {kg}^{- 1}

^{\circ} C^{- 1}

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