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We have a 45g sample of water at it's boiling point, how many joules of energy will be needed to completely evaporate the sample?

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Answer:

The amount of energy needed to raise the temperature of 1 g of water by 1 degree Celcius or (1 Kelvin) equals 1 calorie. And 1 calorie equals 4,184 Joule. So if you want to raise the temperature of 1 litre of water (1 litre = 1000g for water) by 80 degrees Celcius, your calculation should look like this: 1000g*4,184J/(g*°C)*80°C= 334 720 Joules

1 kWh = 3 600 000 Joules, because 1 watt of power during 1 second equals 1 Joule. So 3600 seconds in 1 hour multiplied with a power of 1000 Watt equals 3 600 000 Joules.

334 720 / 3 600 000 = 0,093 kWh of energy to heat 1 kg of water from 20 °C tot 100 °C

If you want to deliver 0,093 kWh in 1 minute instead of 1 hour, the required power output during that minute would be 0,093*60 = 5,58 kW or 5580 Watt.

This also passes the common sense test since a regulator water boiler (for coffee etc.) outputs around 1500 to 2000 watt and using such as device to cook a full litre of water takes more than 1 one minute, more like 3 to 4 minutes which is to be expected if you go through the math.

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