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
- 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.
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.
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