how to work out how much energy is needed to heat water?
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Calculate the kilowatt-hours (kWh) required to heat the water using the following formula: Pt = (4.2 × L × T ) ÷ 3600. Pt is the power used to heat the water, in kWh. L is the number of liters of water that is being heated and T is the difference in temperature from what you started with, listed in degrees Celsius.
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A typical water heater will use around 4000 watts. Click calculate to find the energy consumption of a water heater using 4000 Watts for 3 hours a day @ $0.10 per kWh.
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So in water it takes 1200 joule's to raise a Kg of water by 1 degree K or in degrees C. If this takes 1 min to boil the power required to boil the water equals 96kilo joules/60 seconds =1600 watts or 1.6 kilo watts.
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The energy required to change water from a liquid to a solid is 333.7 kJ/kg while the energy required to boil water is 2257 kJ/kg.
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['Amount of energy released or absorbed is calculated. q = m × Cg × ΔT. q = amount of energy released or absorbed. ... ', 'calculate moles of solute. n = m ÷ M. ... ', 'Amount of energy (heat) released or absorbed per mole of solute is calculated. ΔHsoln = q ÷ n.']
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The heat absorbed is calculated by using the specific heat of water and the equation ΔH=cp×m×ΔT. 4. Water is vaporized to steam at 100oC. The heat absorbed is calculated by multiplying the moles of water by the molar heat of vaporization.
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['Calculate the watts each device uses per day.', 'Convert watts to kilowatts. There are 1000 watts in one kilowatt.', 'Determine the kilowatts an appliance uses per month.', 'Figure out the cost. (Some multiplication required.)']
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Rule of Thumb Estimate of System Volume That is, for closed hot water heating systems that are heated by a hot water boiler, the estimated total volume of the system is determined by multiplying the boiler horsepower by 10. For example, for a 50 hp boiler, the estimated total system volume would be 50 X 10 = 500 USG.
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The equation for calculating heat energy is q=mCpΔT, where q is the heat variable, m is the mass of the object, Cp is the specific heat constant and ΔT is the temperature change.
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In this case, the mass is 2.0g, the specific heat capacity of water is 4.18J/g/K, and the change in temperature is 5.0°C=5K , therefore the energy needed to raise it is: 5×2×4.18=41.8J .
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['2H2 + O2 -> 2H2O. This means that for every 2 molecules of Hydrogen gas, one molecule of Oxygen is needed to make 2 molecules of water. ... ', 'Enthalpy of reaction = (Enthalpy of products) - (Enthalpy of reactants) ... ', 'Enthalpy= -285.83-(0+0)', 'This means Enthalpy of formation of water=-285.83kJ/mol.']
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Most central heating pumps that can be found in a domestic boiler will use around 60 watts of energy, though this could vary depending on the size and speed of your pump. This is less than the amount of energy that's required to boil a kettle.
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That is, water has a high heat of vaporization, the amount of energy needed to change one gram of a liquid substance to a gas at constant temperature. Water's heat of vaporization is around 540 cal/g at 100 °C, water's boiling point.
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So, the total energy required to heat this cup of tea/coffee is: 4.184 * 250 * 79 = 82,634 joules. In other words, it takes more than 82 kJ to heat a cup of water to boiling point!! We usually only need between 1 and 10 joules to heat blood/hair sufficiently to destroy their proteins.
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One of water's most significant properties is that it takes a lot of heat to it to make it get hot. Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.
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A 50-gallon hot water heater with 5,500-watt elements set to 120 degrees takes about 1 hour and 20 minutes to heat water coming in to the unit at 60 degrees.
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Add the change in temperature to your substance's original temperature to find its final heat. For example, if your water was initially at 24 degrees Celsius, its final temperature would be: 24 + 6, or 30 degrees Celsius.