250 g of water at 30° C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and it's contents to 5° C. Given specific latent heat of fusion of ice = 336 x 103 J kg-1, specific heat capacity of copper = 400 J kg-1 K-1, specific heat capacity of water is 4200 J kg-1 K-1.
Given,
mcopper = 50 g
mwater = 250 g
Final temperature = 5° C.
Let mass of ice required be mi.
Heat energy gained by ice at 0° C to convert into water at 0° C
= mi × L
= mi x 336 J
Heat energy gained by (m) g of water at 0° C to rise it's temperature to 5° C
= m x c x change in temperature
= mi x c x (5 - 0) = mi × 4.2 × 5
= 21 x mi
Heat energy lost by water at 30° C in cooling to 5° C
= m x c x change in temperature
= 250 × 4.2 × (30 - 5)
= 250 × 4.2 × 25
= 26250 J
Heat energy lost by vessel at 30° C to cool down to 5° C =
= m x c x change in temperature
= 50 × 0.4 × (30 - 5)
= 50 × 0.4 × 25
= 500 J
If there is no loss of heat,
Heat energy gained = Heat energy lost
Substituting the values in the relation above we get,
Hence, required mass of ice = 74.93 g