If the heat source is 90 degrees, even if the air pressure is low and the boiling point of water is only 90 degrees, the water will not boil. There are two conditions for boiling: reaching boiling point and continuing to boil. the heat. Without a temperature difference, heat cannot be conducted, so it does not boil. You will learn it in college. What is the heating temperature? Can you explain to me, I don't understand?
How many kilowatt hours of electricity does it take to heat 30 liters of water from zero to 90 degrees?
If you are talking about raising the water to 90 degrees Celsius, then you need to use a 105 kilowatt electric heater, excluding losses. From W=Q, we obtain Pt=cm variable t, P ·. 3600s=4.2*10^3J/(kg·℃)*1m^3*1*10^3kg/m^3*90℃, so P=1.05*10^5W=105kW.
V water = 30 liters = 30 liters = 3*10^(-2) cubic meters, T initial = 0 degrees Celsius, T end = 90 degrees Celsius, ρ water = 1*10^3 kilograms/mcubic tres
< p >C water = 4.2*10^3 Joules/(kilogram * degrees Celsius)If heat loss is not included, all electrical energy is converted into thermal energy and absorbed by water, then
E electricity = Q Water absorption
That is, the electrical energy required is E electricity = C water * M water * (T end - T start) = C water * ρ water * V water * (T end - T start)
Obtain electricity E=4.2* 10^3*1*10^3*3*10^(-2)*(90-0) Joule=1.134*10^ 7 Joules
Because 1 kilowatt hour = 1 kilowatt hour = 1,000 watts * 3,600 seconds = 3.6 * 10^6 Joules
So the electrical energy required is E electricity = (1.134* 10^7)/(3.6*10^6 ) kWh = 3.15 kWh