The temperature of $170$ g of water at $50^{o}$ C is lowered to $5^{o}$ C by adding certain amount of ice to it. Find the mass of ice added. Given: Specific heat capacity of water $= 4200$ J $k g^{- 1}$ $^{0} C^{- 1}$ and Specific latent heat of ice $= 336000$ J $k g^{- 1}$ .

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Solution

Let the mass of ice to be added be $m_{i}$ .
Given : $S_{w} = 4200 J / k g^{o} C$ $L = 336000 J / k g$
Mass of water $m_{w} = 170 g = 0.17 k g$
Change in temperature of water $Δ T = 50 - 5 = 45^{o} C$
Heat energy released by water in lowering its temperature is absorbed by ice to melt.
Using $m_{i} L = m_{w} S_{w} Δ T$
$∴$ $m_{i} \times 336000 = 0.17 \times 4200 \times 45$
$⟹ m = 0.0956 k g = 95.6 g$

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