What mass of ice at 0°C added to 2.1 kg water, will cool it down from 75°C to 25°C? Given Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹, Specific latent heat of ice = 336 J g⁻¹.
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Given,
For Water
Mass of water (mw) = 2.1 kg = 2100 g
Initial temperature (Ti) = 75°C
Final temperature (Tf) = 25°C
Specific heat capacity of water (cw) = 4.2 J g⁻¹ °C⁻¹
For ice
Initial temperature (Ti) = 0°C
Final temperature (Tf) = 25°C
Specific latent heat of fusion (Li) = 336 J g⁻¹
Let,
Mass of ice = (mi)
Now,
Heat lost by water = mwcw△t
= 2100 × 4.2 × (75 - 25)
= 2100 × 4.2 × 50 = 441000 J = 441 kJ
Heat gained by ice = miLi + micw△t
= mi × 336 + mi × 4.2 × (25-0)
= 336mi + mi × 4.2 × 25
= 336mi + 105mi = 441mi
By principle of calorimetry,
Heat lost by water = Heat gained by ice
⟹ 441000 = 441 mi
⟹ mi = $\frac{441000}{441}$ = 1000 g = 1 kg
So, the mass of required ice is 1 kg.