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.

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