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3 February, 03:24

How much thermal energy is added to 10.0 g of ice at - 20.0°C to convert it to water vapor at 120.0°C?

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  1. 3 February, 04:24
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    7479 cal.

    31262.2 joules

    Explanation:

    This is a calorimetry problem where water in its three states changes from ice to vapor.

    We must use, the calorimetry formula and the formula for latent heat.

    Q = m. C. ΔT

    Q = Clat. m

    First of all, let's determine the heat for ice, before it melts.

    10 g. 0.5 cal/g°C (0° - (-20°C) = 100 cal

    Now, the ice has melted.

    Q = Clat heat of fusion. 10 g

    Q = 79.7 cal/g. 10 g → 797 cal

    We have water at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

    Q = m. C. ΔT

    Q = 10 g. 1 cal/g°C (100°C - 0°C) → 1000 cal

    Water is ready now, to become vapor so let's determine the heat.

    Q = Clat heat of vaporization. m

    Q = 539.4 cal/g. 10 g → 5394 cal

    Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

    Q = m. C. ΔT

    Q = 10 g. 0.470 cal/g°C. (120°C - 100°C) → 94 cal

    Now, we have to sum all the heat that was added in all the process.

    100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal = 7479 cal.

    We can convert this unit to joules, which is more acceptable for energy terms.

    1 cal is 4.18 Joules.

    Then, 7479 cal are (7479. 4.18) = 31262.2 joules
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