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2. 650g of water at 100 ∘ C is boiled continuously. If the latent heat of vaporisation of water is 2260000J/kg, how much energy is needed to boil the water? Assume that no temperature change occurs.Give your answer to three significant figures.

Question

  1. 650g of water at 100 ∘ C is boiled continuously. If the latent heat of vaporisation of water is 2260000J/kg, how much energy is needed to boil the water? Assume that no temperature change occurs.Give your answer to three significant figures.
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Solution

First, we need to convert the mass of the water from grams to kilograms because the latent heat of vaporisation is given in J/kg.

650g = 0.65kg

The energy (Q) needed to change the state of a substance (in this case, water) can be calculated using the formula:

Q = m * L

where: m = mass of the substance (in kg) L = latent heat of the substance (in J/kg)

Substituting the given values into the formula:

Q = 0.65kg * 2260000J/kg = 1470000J

So, the energy needed to boil the 650g of water is 1470000J or 1.47 * 10^6 J to three significant figures.

This problem has been solved

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