Any number of joules, no matter how small, will raise the temperature
of the water. The total number required in order to accomplish the job
depends on the final temperature you want to see. The higher that is,
the more energy it will take to reach it.
The energy required to boil 100 ml of water at room temperature (20°C) to boiling point (100°C) is about 4200 joules. This is the amount of energy needed to raise the temperature of water by 1°C per gram.
2830 g of water raised through 50 degrees C would use 2830 x 50 calories. But then to boil the water away to steam completely requires another 550 calories per gram, which is 2830 x 550 calories. To convert to Joules, use 4.2 Joules per calorie.
If the water is not already at boiling temperature, then you will need equations 1 and 2. If the water is already at boiling temperature, you will only need equation 2.1. Q = m X C X ΔTThis equation is used to calculate how much energy is required to change the temperature of a given object, of given mass, by a given number of degrees.Q = the total amount of energy required, in joulesm = mass, in grams, of the object being heated (in this case, the water)C = the specific heat of the object (for water, 4.186)ΔT = the total change in temperature2. Q = 2.257 joules X mThe 2.257 in this equation is the heat of vaporization of water: that is, the amount of energy, required per gram of water, to boil water: 2.257 joules per gram. If you were using this equation for a different substance, you would have to look up its heat of vaporization, and substitute it in this equation.Q = the amount of energy required, in joulesm = mass, in grams, of the waterFor example, suppose you were asked to calculate how much energy it would take to boil 256 grams of water which is currently at 40 degrees Celsius. We know that the boiling temperature of water is 100 degrees Celsius; therefore the change in temperature, ΔT, is 100 - 40, which equals 60. Calculate as follows:Q = 256 grams X 4.186 X 60°CQ = 64296.96 joulesThis is how much energy it will take to raise the temperature of the water from 40°C to 100°C. Now calculate how much energy it will take to boil the water once it reaches 100°C:Q = 2.257 joules X 256 gramsQ = 577.792 joulesWe now take the energy required to raise the temperature of the water from 40°C to 100°C and add it to the energy required to boil the water:64296.96 joules + 577.792 joules = 64874.752 joulesConvert to kilojoules:64875.752 joules / 1000 = 64.875752 kilojoulesRound to 64.88 kilojoules.If you are required to express your answer in scientific notation, then express it as6.488 x 103 kilojoules.
No, the amount of heat required to boil 1kg of water is much higher than the amount of heat required to melt 1kg of ice. Boiling water requires additional heat to overcome the latent heat of vaporization, while melting ice only requires heat to overcome the latent heat of fusion.
q (heat energy in Joules) = mass * specific heat * change in temp 1st problem: q = (100 g H2O)(4.180 J/gC)(100 C - 50 C) = 20900 Joules ---------------------- 2nd problem: q = (100 g H2O)(4.180 J/gC)(70 C - 60 C) = 4180 Joules --------------------- As you can see from 50 C to 100 C takes much more heat energy as one would intuitively think, 20900 J/4180 J = 5 times as much energy.
The heat required to boil water can be calculated by multiplying the mass of water (21.1 g) by the specific heat capacity of water (4.18 J/g°C) and the temperature change (100°C - initial temperature). This calculation results in 8.82 kJ or 8820 J of energy needed to boil 21.1 g of water at 100°C.
The energy required to boil 100 ml of water at room temperature (20°C) to boiling point (100°C) is about 4200 joules. This is the amount of energy needed to raise the temperature of water by 1°C per gram.
To determine the volume of water that can be boiled using 3.0 kg of energy, we first convert the energy from kilograms to joules. Assuming the energy is in kilograms of mass equivalent (using E=mc²), 3.0 kg of energy is equivalent to about 2.7 x 10^16 joules. The energy required to boil water is approximately 2,260 joules per gram (the latent heat of vaporization). Thus, dividing the total energy by the energy needed to boil water gives you about 11,900,000 grams, or approximately 11,900 liters of water.
If by "boil" you mean have it all evaporate, that takes MUCH more energy. For example, to increase the temperature of one gram of water from 20 to 100 degrees Celsius, you need 4.2 joules/gram/degree times 80 degrees = about 336 joules; then, to evaporate all the water, you need an additional 2257 joules.
That's going to depend on what temperature the water starts from.
1 kilowatt = 1000 joules per second, so it will take 480 seconds. (8 minutes).
100 degrees Celsius
To bring 1 liter of water to a boil, you would need approximately 0.24 kilograms of wood. This is based on the energy content of wood, which is around 16-20 megajoules per kilogram. The specific amount of wood required can vary depending on factors such as the type of wood, its moisture content, and the efficiency of the burning process.
With enough insulation you can in theory boil any amount of water with any amount of energy, given enough time.The key is to supply more energy to the water than it looses.It takes 2260 joules to boil 1 cc of water, joules can be described at watt-seconds.One pint of water is approximately 473 cc, so that will requite 1,068,980 Joules to boil.Given perfect insulation; it would take 17816 seconds / 297 minutes / 5 hours to boil the water with a 60 watt heating element.Assuming that you do not use perfect insulation and given the fact that a peltier element won't give out all 60 watts on the one side, i'd have to say;No you can't boil a pint of water on a 60 watt peltier unit.
You need o know the initial temperature of water:Heat = specific heat x difference of temperature x mass of water
2830 g of water raised through 50 degrees C would use 2830 x 50 calories. But then to boil the water away to steam completely requires another 550 calories per gram, which is 2830 x 550 calories. To convert to Joules, use 4.2 Joules per calorie.
It would take 4.5 minutes for a kettle to boil 1.5 liters of water since it takes 3 minutes to boil 1 liter.