Answer:
mass of the ice that will be added is 0.29 kg
Explanation:
Given;
mass of water, M₁ = 1.16 kg
temperature of water, T₁ = 34 °C
final temperature of water-ice system, T₂ = 11 °C
latent heat of fusion for water, L = 33.5 × 10⁴ J/kg
specific heat capacity of water, c = 4186 J/(kg·C°)
Apply the principle of conservation energy;
Heat lost by water when it is cooled from 34 °C to 11 °C must be equal to heat gained by the ice when it melts.
M₁cΔθ = M₀L + M₀cΔθ
where;
M₁ is mass of water
M₀ is mass of ice
c is specific heat capacity of water
Δθ is change in temperature
L is latent heat of fusion of ice
M₁cΔθ = M₀L + M₀cΔθ
M₁cΔθ = M₀(L + cΔθ)
Mc (T₁ - T₂) = M₀ [L + c(T₂ - 0)]
1.16 x 4186 x (34 - 11) = M₀ [335000 + 4186 x 11]
111682.48 = M₀ (335000 + 46046)
111682.48 = 381046M₀
M₀ = (111682.48) / (381046)
M₀ = 0.29 kg
Therefore, mass of the ice that will be added is 0.29 kg
