How much kilowatt hour electricity is needed to heat 400 kg of water with a heater with a power of 3

Q: How much kilowatt-hour does it take to heat 400kg of water with a 350w power heater.

Answer: To heat 400 kg of water with a heater with a power of 350w, for every 1 degree of rise, you need to use to kWh (kWh to kWh). It takes up to hours.

1. 1 gram of water warms up by 1 degree, which requires joule energy.

1 kg (1000 g) of water warms up to 1 degree and requires 4200 joules of energy (joules * 4200 joules per 1000 g).

100 kg of water warms up to 1 degree and requires 420,000 joules of energy (4,200 joules * 100 kg 420,000 joules).

400 kg of water warms up to 1 degree, which requires 1,680,000 joules of energy (420,000 joules * 4 hundred kg 1,680,000 joules).

2. 350W, 350 joules of energy in 1 second.

350W, 1 hour (3600 seconds) produces 1,260,000 joules of energy (350 joules * 3,600 seconds 1,260,000 joules).

3. 350w raises 400 kg of water by 1 degree, which takes hours (1680000 1260000.

At 80% ( efficiency, it takes hours (hours hours).

Fourth, 1000W, 1 second to produce 1000 joules of energy.

1 kWh of electricity 1000W works for 1 hour (3600 seconds), which is 3600000 joules (1000 joules * 360000 seconds 3600000 joules).

5. 350W work for 1 hour, using electricity (350W hours, 1000W hours, or 1260000 joules 3600000 joules per hour).

350W working hours, 400 kg of water is heated by 1 degree, and electricity is used (kWh * kWh per hour).

6. If the heat is fast, there is no poor insulation facilities, and the thermal efficiency is 80% (calculated, 350W will heat up 400 kg of water by 1 degree, then electricity is required: kWh (kWh).

or kilowatt-hours * hours of electricity.

7. 350w, boiling 400 kg of water, the power is too small. Now the water temperature is about 20 degrees, and the bath water needs to be heated to 50 degrees Celsius, and it needs to be heated by 30 degrees Celsius for 40 to 50 hours. Burn here, and the heat here is dissipated and lost a lot.

8. It should be 3500w.

Or 350w to burn 40 kg of water, it's pretty much the same.

The specific heat capacity of the water is required, the initial temperature of the water.

If thermal efficiency is not considered, it is sufficient to use energy conservation.

1 degree = 3600000J

No one can give the right answer to this question.

Because no one knows the working efficiency of a 350W heater.

Summary. Hello, it's a pleasure to serve you! It takes about 2 hours to heat 380 kg of water from 30 degrees to 100 degrees with a 20 kW heater.

Hello, it's a pleasure to serve you! It takes about 2 hours to heat 380 kg of water from 30 degrees to 100 degrees with a 20 kW heater.

Q suction = paisan 100-30) = 111720 kilojoules 111720 = 22xtt = 111720 22t = seconds t = hours, so theoretically about 2 hours can be from 30 degrees to 100 degrees from the dust front heat.

Summary. Hello, glad to answer for you! Pro-40 cubic water with a 200kw heater from 10 to 40 degrees takes hours, because with a heater with a power of 200kw to heat 40 cubic water, according to the second law of thermodynamics, the heating time is equal to the required heat divided by the power, ie:

t=q/p=40m³×4200j/kg×30k/m³/200kw=。

How long does it take to heat 40 cubic meters of water from 10 to 40 degrees with a 200kw heater.

Good. Detailed calculations.

Heater efficiency 95

Hello, glad to answer for you! It takes hours to heat 40 cubic meters of water from 10 to 40 degrees with a 200kw heater, because it takes hours to heat 40 cubic meters of water with a heater with a power of 200kw, according to the second law of thermodynamics, the heating time lead tan is equal to the heat required by the Huai Yutong divided by the power, that is: t=q p=40m 4200j kg 30km 200kw=.

That's about 3 hours and 10 minutes.

Yes dear. I'm an electromagnetic heater.

Efficiency 95 helps you do the math.

If you heat 9,540 cubic meters of water from 10 to 40 degrees Celsius with a 200kw heater, it will take about an hour. In addition, the final heating time is also affected by the temperature and temperature difference between the water and the water of the plum limb ear, the temperature of the outdoor environment, the wind speed and humidity, and the thermal efficiency of the heater.

400 watts of dong eggplant, with an hour, with a blind tremor wide 1 kilowatt-hour of electricity.

That is: 1,000 grinding watts x hours = degrees.

400 kW. x

There is still a time limit for your question: how long can I heat the water to the temperature set in the finger band, otherwise I can use as much power as I want, because the time limit for kerosene is high, and it will take longer if the power is high.

Q suction = Gyeongsan if heat loss is not counted: Q suction = Q discharge = pt

So: p=q sucks t

Pro, to calculate the size of the electric heating rod required by Yuxiang, the following factors need to be considered:1The quality of the water town mountain fight:

The mass of 500 liters of water is 500 kilograms. 2.Heating temperature difference:

Assuming a heating time of 1 hour. 4.Heat loss:

The tank will have a certain amount of heat loss, but it is assumed that this heat loss is negligible. Based on these factors, the required electric heating rod size can be calculated using the following formula: Heat = Mass of Water Temperature Difference Specific heat capacity where the specific heat capacity is the specific heat capacity of water and is j g.

Heat = 500 kg 15 Jg = 31,350,000 J To convert heat into power, you can use the following formula: Power = Heat Heating time power = 31,350,000 J 3600 seconds = 8,708 watts Therefore, an electric heating rod with a power of 8,708 watts is required to heat a 500 liter tank from 30 degrees to 45 degrees. However, it should be noted that the appropriate electric heating rod should be selected for actual use, and should not exceed the load capacity of the circuit and socket to ensure safety.

To calculate the work-to-material return rate of the electric heating rod required to heat the tank, it is necessary to consider the volume of the tank, the quality of the water, the initial and target temperatures, and the heating time. Suppose the volume of the tank is 500 liters, the mass of water is 500 kg, the initial temperature is 30 degrees, the target temperature is 45 degrees, and the heating time is 1 hour. First of all, you need to calculate the heat capacity of the water during heating.

The heat capacity of water is j (g.).) is the amount of heat required to raise by 1 degree per gram of water. Therefore, the amount of heat required to heat 500 kg of water from 30 to 45 degrees is:

q = m * c * t where m is the mass of water, c is the heat capacity of water, and δt is the change of temperature deficit. q = 500 * 45 - 30) = 31,380 j, the next step is to calculate the power required for the electric heating rod. The power of the electric heating rod can be calculated by the following formula:

p = q t where p is the power, q is the heat, and t is the heating time. p = 31,380 3600 = kW Therefore, the electric rod power required to heat a 500 liter tank from 30 to 45 degrees is kilowatts. Actually, this is only a rough estimate, as there are other factors to consider, such as the efficiency of the electric heating rod, the material used to heat the tank, etc.

3 tons of water 50kw heater from 20 degrees to 45 degrees how long does it take, how many degrees of electricity.

The heat capacity of the water is that Sun Zhi heats a ton of water from 20 degrees to 50 degrees, and the heat energy required is 1000 * 1kwh = 3600kj electricity consumption = 126000 3600 = 35 degrees.