104 g of water at 30°C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added.
[Specific heat capacity of water = 4.2 J g^-1 °C^-1; Specific latent heat of fusion of ice = 336 J g^-1; Specific heat capacity of copper = 0.4 J g^-1 °C^-1 ]

Given,

Mass of water = 104 g

Temperature of water = 30°C

Final temperature = 10°C

Mass of calorimeter = 42 g

Temperature of ice = 0°C

The heat given by water = m x s x t

Substituting the values we get,

The heat given by water = 104 x 4.2 x (30 - 10) = 8736 J

The heat given by calorimeter = m x s x t

Substituting the values we get,

The heat given by calorimeter = 42 x 0.4 x (30 - 10) = 336 J

Let mass of ice added be m gm

Heat taken by ice to melt = mL = 336m J

Heat taken by ice-water to reach the mixture temperature = mst = m x 4.2 x (10 - 0) = 42m J

Therefore, from principle of calorimetry, heat taken = heat given.

i.e., 336 m + 42 m = 8736 + 336

Therefore, m = $\dfrac{9072}{378}$ = 24 g

Hence, mass of ice added = 24 g