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AURORKA [14]
3 years ago
10

Consider a hot-air balloon. The deflated balloon, gondola, and two passengers have a combined mass of 315 kg. When inflated, the

balloon contains 845 m^3 of hot air. Find the temperature (∘C) of the hot air required to lift the balloon off the ground. The air outside the balloon has a temperature of 20.2∘C and a pressure of 0.916×105 Pa. The pressure of the hot air inside the balloon is the same as the pressure of the air outside the balloon. The molar mass of air is 29 g/mol. Ignore the buoyant force on the gondola and the passengers. Hint: First find the density of the air outside the balloon. Then use Archimedes' principle to find the density required for the air inside the balloon. Then find the temperature of the air inside the balloon.
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0

This will help you.

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The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

<h3>How to calculate the time</h3>

Use the formula:

Power × time = mass × specific heat

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Specific latent heat of vaporization = 4000000 J/ Kg

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Substitute the values into the formula

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2500 × time = 1. 5 × 4000000

Make 'time' the subject

time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds

Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

Learn more about specific latent heat of vaporization:

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