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3 March, 09:07

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.

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  1. 3 March, 10:20
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    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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