A vessel of mass $100 g$ (S.H.C. $= 0.2 cal g^{- 1} {{}^{\circ}C}^{- 1}$ ] contains $500 g$ of water at $37 ° C$ . Calculate the amount of ice, which should be added to the vessel, so that the final temperature is $17 ° C$ .
[S.H.C. of water $= 1 cal g^{- 1} {{}^{\circ}C}^{- 1}$ and S.L.H. of ice $= 80 cal g^{- 1}$ ]

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Solution

Step 1: Given data:
SHC of water $= c_{w} = 1 cal g^{- 1} {{}^{\circ}C}^{- 1}$
SHC of vessel $= c_{v} = 0.2 cal g^{- 1} {{}^{\circ}C}^{- 1}$
SLH of ice $= L = 80cal. g^{- 1}$
Mass of vessel $= m_{1} = 100 g$
Mass of water $= m_{2} = 500 g$
Initial temperature $= T_{2} = 37 ° C$
Final temperature $= T = 17 ° C$
Step 2: Calculating the mass of ice that needs to be added
Let the mass of ice added be $M g r a m s .$
We know, Heat gained by ice $=$ Heat lost by [vessel $+$ water]
$[ML + M c_{w} \times (T - 0) = [m_{1} c_{v} + m_{2} c_{w}] [T_{2} - T]$
On substituting the values we will get,
$[ML + M c_{w} \times (17 - 0) = [m_{1} c_{v} + m_{2} c_{w}] [37 - 17] M [80 + (1 \times 17)] = [(100 \times \frac{2}{10}) + (500 \times 1)] \times 20 M [97] = [520] \times 20 M = \frac{520 \times 20}{97} = 107.2 g$
Hence, the amount of ice to be added is $107.2 g$

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