$500 g$ of water at $60 ° C$ is contained in a vessel of negligible heat capacity. Into this water is added $400 g$ of ice at $0 ° C$ . Calculate the amount of ice which does not melt.
[Take $S . H . C$ of water $= 4.2 J g^{- 1} {{}^{\circ}C}^{- 1}$ and $S . L . H$ of ice $= 336 {Jg}^{- 1}$ ]

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Solution

Step 1: Given data,
Mass of water $= 500 g$ at $60 ° C$
Mass of ice $= 400 g$ at $0 ° C$
SHC of water $= c = 4.2 J g^{- 1} {{}^{\circ}C}^{- 1}$
SLH of ice $= L = 336 {Jg}^{- 1}$
Let $T$ be the temperature of the water $= 60 ° C$ .
Step 2: Finding the amount of ice which does not melt.
Heat lost by $500 g$ water at $60 ° C$ to cool to $0 ° C = m c (T - 0)$
$= 500 \times 4.2 \times (60 - 0) J = 500 \times 4.2 \times 60 J$

This heat loss is used in melting $m$ gram of ice at $0 ° C$ , so
$m L = 500 \times \frac{42}{10} \times 60 m \times 336 = 3000 \times 42 m = \frac{3000 \times 42}{336} = 375 g$
Amount of ice which does not melt $= 400 - 375 = 25 g$ .
Hence, the 25g of ice will not melt.

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500g of water at 60°C is contained in a vessel of negligible heat capacity. Into this water is added 400g of ice at 0°C. Calculate the amount of ice which does not melt.[Take S.H.C of water =4.2Jg-1C∘-1 and S.L.H of ice =336Jg-1 ]

$500 g$ of water at $60 ° C$ is contained in a vessel of negligible heat capacity. Into this water is added $400 g$ of ice at $0 ° C$ . Calculate the amount of ice which does not melt.
[Take $S . H . C$ of water $= 4.2 J g^{- 1} {{}^{\circ}C}^{- 1}$ and $S . L . H$ of ice $= 336 {Jg}^{- 1}$ ]