ALL ABSURD ANSWERS WILL BE REPORTED AND REMOVED! 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?

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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