4186 Joules per liter per deg C. Not clear if we are raising the temperature BY 135 deg or TO 135 deg. So the answer is 4186 x 100 x rise in temperature. (Joules).
Kilowatts is a unit of energy rate, while the temperature required to raise a specific volume of water by a specific amount of degrees is a unit of energy, not energy rate. The question cannot, therefore, be answered as stated. Please restate the question.
To calculate the energy required to heat the geyser, you can use the formula: Energy (kWh) = Volume (liters) x Temperature rise (degrees Celsius) x 4.18 (specific heat capacity of water) / 3600 Plugging in the values, the calculation would be: 150 liters x 65 degrees x 4.18 / 3600 = approximately 30.45 kWh.
1 litre = 1000 millilitres. You now have all the information required to answer this and similar questions.
If it's 23 degrees Celsius the amount of heat energy for that is 100 x 23 kilocalories, and it is independent of the time. 1 kilocalorie is 4200 Joules so the energy could also be expressed as 9.66 Megajoules. 3.6 MJ is 1 kilowatt-hour so the above energy is also 2.683 kilowatt-hours.
2 liters of hot water will have more heat energy than 1 liter of hot water because there is a greater quantity of water to hold heat energy. The total heat energy in a system is directly proportional to the mass of the substance and its temperature.
Kilowatts is a unit of energy rate, while the temperature required to raise a specific volume of water by a specific amount of degrees is a unit of energy, not energy rate. The question cannot, therefore, be answered as stated. Please restate the question.
To calculate the energy required to heat the geyser, you can use the formula: Energy (kWh) = Volume (liters) x Temperature rise (degrees Celsius) x 4.18 (specific heat capacity of water) / 3600 Plugging in the values, the calculation would be: 150 liters x 65 degrees x 4.18 / 3600 = approximately 30.45 kWh.
To calculate the energy needed to heat water, you can use the formula: energy = mass of water (kg) x specific heat capacity of water (4.18 J/g°C) x change in temperature (°C). Converting 50 liters to 50 kg, the energy needed would be: 50 kg x 4.18 J/g°C x 50 °C = 10,450 J, or 10.45 kJ.
2,000 liters of water (at 4 degrees Centigrade) weighs 2,000 kilograms. This follows from the definitions of the units involved.
Liters liquid 1000ml/1L g/ml mol/g Hfusion
39.25 liters.
Assuming the water begins at room temperature of 25 degrees Celsius: about 12.5 liters.
That depends on the model of the transformer. I have seen transformers that required 7500 liters and some that needed 2,5 liters.
1 imperial gallon is about 4.5 liters (to an absurd degree of detail = 4.54609188 litres). To convert a value in imperial gallons to litres, multiply by 0.22 or 0.219969248299.1 US gallon is about 3.8 liters (to an absurd degree of detail = 3.78541178 liters)(isn't it stupid to have two different gallons?)
Liters liquid 1000ml/1L g/ml mol/g Hfusion
More information is required to answer this question. 3.5 what?
2