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# The specific heat of water $=4200\text{\hspace{0.17em}\hspace{0.17em}}J\text{\hspace{0.17em}\hspace{0.17em}}k{g}^{-1}\text{\hspace{0.17em}}{K}^{-1}$  and the latent heat of ice  $=3.4×{10}^{5}\text{\hspace{0.17em}\hspace{0.17em}}J\text{\hspace{0.17em}\hspace{0.17em}}k{g}^{-1}$. 100 grams of ice at  $0°C$ is placed in 200 g of water at  $25°C$. The amount of ice that will melt as the temperature of water reaches  $0°C$ is close to (in grams) :

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By Expert Faculty of Sri Chaitanya
a
61.7
b
64.6
c
63.8
d
69.3
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detailed solution

Correct option is A

Heat extracted out of system  =0.2420025=21000 JSo amount of ice melt  =QL=210003.4×105  kg=2103.4 g=61.76  gm

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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