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4 February, 10:26

An 8.5 g ice cube is placed into 255 g of water. Calculate the temperature change in the water upon the complete melting of the ice. Assume that all of the energy required to melt the ice comes from the water.

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  1. 4 February, 10:36
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    Change in temperature of water = 2.67 °C

    Explanation:

    Heat gained by ice = heat lost by water

    Q₁ = Q₂ ... Equation 1

    Where Q₁ = Latent heat of ice, Q₂ = heat lost by water

    Q₁ = lm₁ ... equation 2

    Q₂ = cm₂ΔT ... Equation 3

    Substituting equation 2 and 3 into equation 1

    lm₁ = cm₂ΔT ... Equation 4

    Making ΔT the subject of formula in equation 4

    ΔT = lm₁/cm₂ ... Equation 5

    Where ΔT = change in temperature of water, l = specific latent heat of ice, m₁ = mass of ice, c = specific heat capacity of water, m₂ = mass of water.

    Given: m₁ = 8.5 g = 0.0085 kg, m₂ = 255 g = 0.255 kg.

    Constant: l = 336000 J/kg, c = 4200 J/kg. K

    Substituting these values into equation 5

    ΔT = (336000*0.0085) / (0.255*4200)

    ΔT = 2856/1071

    ΔT = 2.67 °C

    Change in temperature of water = 2.67 °C
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