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23 September, 01:07

What mass of LP gas is necessary to heat 1.7 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C) ? Assume that during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

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  1. 23 September, 01:34
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    72.25 g

    Explanation:

    mass of 1.7 L water = 1.7 x 10⁻³ x 10³ kg

    = 1.7 kg

    heat required to raise its temperature from 25 degree to 100 degree

    = mass x specific heat x rise in temperature

    = 1.7 x 4.18 x 10³ x 75 J

    = 532.95 kJ

    Now calorific value of LP gas = 46.1 x 10⁶J / kg

    Let required mass of LP gas be m kg

    heat evolved from this amount of gas

    = 46.1 x 10⁶ m

    Heat utilized in warming water

    = 46.1 x 10⁶ m x. 16 J

    So

    46.1 x 10⁶ m x. 16 = 532.95 x 10³

    7376 x 10³ m = 532.95 x 10³

    m = 532.95 / 7376 kg

    = 72.25 g
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