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2.0g of water at 100 ∘ C is boiled without the temperature changing. The latent heat of vaporisation of water is 2260000J/kg. How much energy is needed to boil the water?

Question

2.0g of water at 100 ∘ C is boiled without the temperature changing. The latent heat of vaporisation of water is 2260000J/kg. How much energy is needed to boil the water?

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Solution

Step 1: Convert the mass of water from grams to kilograms. 1 kg = 1000 g, so 2.0 g = 2.0/1000 = 0.002 kg.

Step 2: Use the formula for heat energy, which is Q = mL, where m is the mass of the substance and L is the latent heat of vaporisation.

Step 3: Substitute the given values into the formula. Q = 0.002 kg * 2260000 J/kg = 4520 J.

So, 4520 Joules of energy is needed to boil the 2.0g of water.

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