The temperature of 1.16 kg of water is 34 °C. To cool the water, ice at 0 °C is added to it. The desired final temperature of the water is 11 °C. The latent heat of fusion for water is 33.5 × 104 J/kg, and the specific heat capacity of water is 4186 J/(kg·C°). Ignoring the container and any heat lost or gained to or from the surroundings, determine how much mass m of ice should be added.

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

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explanation:

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15.3 cos 65 degrees trig