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6 August, 22:32

Suppose 9.30 3 105 J of energy are transferred to 2.00 kg of ice at 0°C. (a) Calculate the energy required to melt all the ice into liquid water. (b) How much energy remains to raise the temperature of the liquid water? (c) Determine the final temperature of the liquid water in Celsius.

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  1. 6 August, 22:47
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    (a) 672000 J

    (b) 258300 J

    (c) 30.75 °C

    Explanation:

    (a) Energy required to melt the ice = Latent heat of fusion of ice

    Latent heat of fusion of ice (Q₁) = lm ... Equation 1

    Where l = specific latent heat of fusion of ice, m = mass of ice

    Given: m = 2.00 kg

    Constant : l = 336000 J/kg

    Substituting these values into equation 1,

    Q₁ = 336000 * 2

    Q₁ = 672000 J.

    Energy required to melt the ice = 672000 J

    (b) The Energy remains to raise the temperature of the liquid water = Total Energy - Energy requires to melt the ice.

    Given: Total Energy = 9.303 * 10⁵ J, = 930300 J

    Energy remain to raise the temperature of the liquid water =

    930300 - 672000

    Total Energy = 258300 J

    Energy remain to raise the temperature of the liquid water = 258300 J.

    (c) : Q = cmΔT ... equation 2

    Where c = specific heat capacity of water, m = mass of water, ΔT = change in temperature, Q = Energy required to change the temperature of liquid water

    Making ΔT the subject of formula in the equation above,

    ΔT = Q/cm ... Equation 3

    Given: m = 2.00 kg, Q = 258300 J

    Constant: C = 4200 J/kg. K

    Substituting these values into equation 3

    ΔT = 258300 / (4200*2)

    ΔT = 258300/8400

    ΔT = 30.75 K

    But ΔT = T₂ - T₁

    Where T₁ initial temperature, T₂ = final Temperature

    T₂ = T₁ + ΔT,

    Given: T₁ = 0 °C

    Therefore,

    T₂ = 30.75 + 0 = 30.75

    T₂ = 30.75 °C

    Final temperature of water = 30.75 °C
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