$10 g$ of ice at $0 ° C$ is added to $10 g$ of water at $80 ° C$ , such that the temperature of mixture is $0 ° C$ . Calculate the sp. latent heat of ice. [S.H.C. of water $= 4.2 {Jg}^{- 1} {{}^{\circ}C}^{- 1}$ ]

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Solution

Step 1: Given data,
Mass of ice $= M = 10 g$
Mass of water $= m = 10 g$
SHC of water $= c_{w} = 4.2 {Jg}^{- 1} {{}^{\circ}C}^{- 1}$
Initial temperature= $= T_{2} = 80 ° C$
Final temperature $= T = 0 ° C$
Step 2 : Finding the specific latent heat capacity of ice.
Let specific latent heat capacity of ice $= L$
We will have, Heat gained by ice = Heat lost by water
$ML = m c_{w} (T_{2} - T)$
On substituting the values, we will get,
$(10 \times L) = 10 \times \frac{42}{10} (80 - 0) L = \frac{10 \times 42 \times 8}{10} = 336 {Jg}^{- 1}$
Hence, specific latent heat capacity of ice is $336 J g^{- 1}$ .

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Q. 10 g of ice at 0°c is added to 10 g of water at 80°c such that the final temp. of the mixture is 0°c.Calculate the sp.latent heat of ice? ( S.H.C. of water = 4.2 J/g/°c)

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Water at $80^{\circ} C$ is poured into a bucket containing $1.5 k g$ of crushed ice at $0^{\circ} C$ , such that all the ice melts and the final temperature recorded is $0^{\circ} C$ . Calculate the amount of hot water added to the ice.[Take the specific heat capacity of water is $4200 J K g^{- 1} K^{- 1}$ and the specific latent heat of ice $= 336 \times 10^{3} j K g^{- 1}$ .]