A vessel of mass $80 g$ (S.H.C. $= 0.8 {Jg}^{- 1} C^{- 1}$ ) contains $250 g$ of water at $35 ° C$ . Calculate the amount of ice at $0 ° C$ , which must be added to it, so that final temperature is $5 ° C$ .
[Sp. latent heat of ice $= 340 {Jg}^{- 1}$ ]

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Solution

Step 1: Given data:
Mass of vessel $= m_{1} = 80 g$
Mass of water $= m_{2} = 250 g$
Initial temperature $= T_{2} = 35 ° C$
Final temperature $= T = 5 ° C$
SHC of vessel $= c_{v} = 0.8 J g^{- 1} K^{- 1}$
SLH of ice $= L = 340 J g^{- 1}$
SHC of water $= c_{w} = = 4.2 {Jg}^{- 1} {{}^{\circ}C}^{- 1}$
Step 2: Finding the amount of ice to be added.
Let the mass of ice that needs to be added be $M g r a m s .$
We know,
Heat gained by ice $=$ Heat lost by (container $+$ water)
$ML + M c_{w} (T - 0) = [m_{1} c_{v} + m_{2} c_{w}] (T_{2} - T)$
On substituting the values we will get,
$M [L + (c_{w} \times 5)] = [(80 \times \frac{8}{10}) + 250 \times \frac{42}{10}] (35 - 5) M [340 + (\frac{42}{10} \times 5)] = (64 + 1050) \times 30 361 M = 33420 M = \frac{33420}{361} M = 92.57 g$
Hence the amount of ice that needs to be added is $92.57 g$

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Q. A vessel of mass 80 g(S.H.C. = 0.8 J/g/°c) contains 250g of water at 35°c.Calculate the amount of ice at 0°c, which must be added to it,so that the final temp. is 5°c? (Sp. latent heat of ice=340J/g)

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A vessel of mass 80g(S.H.C. =0.8Jg-1C-1 ) contains 250g of water at 35°C. Calculate the amount of ice at 0°C, which must be added to it, so that final temperature is 5°C.[Sp. latent heat of ice =340Jg-1 ]

A vessel of mass $80 g$ (S.H.C. $= 0.8 {Jg}^{- 1} C^{- 1}$ ) contains $250 g$ of water at $35 ° C$ . Calculate the amount of ice at $0 ° C$ , which must be added to it, so that final temperature is $5 ° C$ .
[Sp. latent heat of ice $= 340 {Jg}^{- 1}$ ]