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27 May, 01:51

A 3.8 kg block of copper at a temperature of 84°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.)

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Answers (2)
  1. 27 May, 03:08
    0
    0.211 kg

    Explanation:

    specific heat capacity of copper = 385 J/Kgk

    heat loss by copper = mcθ = 3.8 * (8 - 84) * 385 = - 111188 J

    heat needed to raise the temperature of water from 0°C to 8°C

    = mcθ = 1.2 kg * 4180 * (8 - 0) = 40128 J

    111188 J - 40128 J = 71060

    71060 = ml

    71060 / 336000 = mass of ice where latent heat of fusion = 3.36 * 10⁵JKg⁻¹

    m = 0.211 kg
  2. 27 May, 05:29
    0
    0.215 kg of ice

    Explanation:

    Mass of copper, m = 3.8 kg

    Specific heat of Copper, Cp = 385 J/kg•K;

    Change in temperature = 84 - 8

    = 76°C

    q = m * Cp * delta T

    = 385 * 3.8 * 76

    = 111188 J

    111188 J = 26.575 kcal

    The energy needed to heat 1.2 kg of water from 0°C to 8°C = q

    q = m * Cp * delta T

    = (1.2 kg) (1 kcal/kg•°C) (8°C)

    = 9.6 kcal

    Therefore,

    delta q = q1 - q2

    = 26.575 - 9.6

    = 16.975 kcal were used to melt the ice.

    Energy needed to melt ice at 0°C = 79 kcal/kg

    Mass = 16.975/79

    = 0.215 kg of ice.

    14 kcal / (79 kcal/kg) = 0.18 kg.
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