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30 May, 19:50

A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice melt, what was the original mass of the block of ice?

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  1. 30 May, 19:55
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    original mass of the block of ice is 38.34 gram

    Explanation:

    Given data

    cup mass = 150 g

    ice temperature = 0°C

    water mass = 210 g

    water temperature = 12°C

    ice melt = 2 gram

    to find out

    solution

    we know here

    specific heat of aluminum is c = 0.900 joule/gram °C

    Specific heat of water C = 4.186 joule/gram °C

    so here temperature difference is dt = 12 - 0 = 12°C

    so here heat lost by water and cup are given by

    heat lost = cup mass * c * dt + water mass * C * dt

    heat lost = 150 * 0.900 * 12 + 210 * 4.186 * 12

    heat lost = 12168.72 J

    so

    mass of ice melt here = heat lost / latent heat of fusion

    here we know latent heat of fusion = 334.88 joule/gram

    so

    mass of ice melt = 12168.72 / 334.88

    mass of ice melt is 36.337554 gram

    so mass of ice is here = mass of ice melt + ice melt

    mass of ice = 36.337554 + 2

    mass of ice = 38.337554 gram

    so original mass of the block of ice is 38.34 gram
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