shut de routes
226,ooo j
The number of calories required will depend on the mass of water which is to be heated.
Kilowatt is a measure of the rate of energy use. It is 1,000 Watts or 1,000 Joules per second. A kilowatt hour is 1,000 Joules per second for 3,600 seconds or 3,600,000 Joules. This means a Kilowatt hour not a rate but a measure of total energy used.
27 Joules
Work is the force times distance, so the answer is 2,850,000 joules.
It's necessary to remove 540 calories from a gram of water in order to freeze it. That's about 2260 joules. The amount of energy used by a freezer to do this depends on the efficiency of the freezer.
The temperature varies from one liquid to another. Every element has a different freezing point and boiling point. You can calculate both for a molecule by using a formula that takes into consideration the different elements that make up the molecule in comparision to percentage that element is of the total mass of the molecule
Approx 4974 Joules.
a change from liquid to solid or solid to liquid will require the use of the formula of Heat of Fusion , q=mHf where m is the mass and Hf = 334J/g, this is standard so you wont have to look for how it came about.In this case the m=5.00g so: q = m Hf q=(5.00g)(334 J/g) , you notice the grams will cancel out. q = 1670 J.
226,ooo j
226,ooo j
The number of calories required will depend on the mass of water which is to be heated.
a change from liquid to solid or solid to liquid will require the use of the formula of Heat of Fusion , q=mHf where m is the mass and Hf = 334J/g, this is standard so you wont have to look for how it came about.In this case the m=5.00g so: q = m Hf q=(5.00g)(334 J/g) , you notice the grams will cancel out. q = 1670 J.
Energy is actually given off in the lowering of temperature. Use the equation Q = mc(change in T). In this case, m = 2.9, c = 4.179 J, and change in T = -12.1 degree C. Now solve for Q, the heat energy. This exothermic process gives off approximately 146.6 J of heat.
Water has a specific heat of 1 calorie per gram per degree. This means that in order to raise 28.4 grams of water by 1 degree Celsius, it takes 28.4 calories. At an increase of 44.6 degrees Celsius, this equates to 1,266.64 calories of heat.
yes kilojoues convert down to joules which is the amount of energy in the body total joules= total energy
Watts = joules / secondYou need to divide the total annual energy consumption by the number of seconds in a year (which is about 31.5 million).