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To raise the temperature of both an equal amount, water would require more energy. In terms of the energy required to raise the temperature: iron = 0.45 joules / gram . kelvin water = 4.2 joules / gram . kelvin This is known as the specific heat capacity of a material
Any number of joules, no matter how small, will raise the temperatureof the water. The total number required in order to accomplish the jobdepends on the final temperature you want to see. The higher that is,the more energy it will take to reach it.
9.6 MJ
shc of graphite = 710 j - kg - k > 350 joules would raise 1 kg by (350 / 710) 0.49295 deg kelvin, so would raise temp. of 33 kg by 0.493 / 33 = 0.014938 kelvin
Assuming the water is liquid, the specific heat is about 4.186 joule/gram·°C, so to heat 46 grams of water would take about 192.556 joules/°C. The specific heat of ice is about 2.100 Joules/g·°C so heating 46 g of frozen water would take about 96.6 joules/°C. The specific heat of steam is about 2.020 Joules/g·°C so heating 46 g of water vapor would take about 92.2 joules/°C.
To raise the temperature of both an equal amount, water would require more energy. In terms of the energy required to raise the temperature: iron = 0.45 joules / gram . kelvin water = 4.2 joules / gram . kelvin This is known as the specific heat capacity of a material
Any number of joules, no matter how small, will raise the temperatureof the water. The total number required in order to accomplish the jobdepends on the final temperature you want to see. The higher that is,the more energy it will take to reach it.
Well, let's see. Water at room temperature has a heat capacity of 4.18 J/g-C, and water also has a density of 1g/mL. If there's one litre of water, there's 1000 g of water. If the change in temperature is 1 C, and there are 1000 g of water, and specific heat capacity's 4.18... Q = mcT Q = (1000g)(4.18J/g-C)(1 C) Q = 4180 J So you need 4180 J of heat. *************************************** The definition of one calorie is as stated in the question. One calorie is equivalent to 4.18 Joules. Not sure where the maths went wrong but just so you know.
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
10ml's of water is equal to 10cm3 of water. 10cm3 of water has a mass of 10g. The specific heat of water is 4.134 J/K. The change in temperature is 1 degree Kelvin. Use Q=mC∆T which means Heat= (Mass)(Specific Heat)(Change in Temperature) Q= (10)(4.134)(1) Q=(10)(4.134) Q=41.34 Joules
The answer is WATER!
You don't decrease the temperature, you raise the water's boiling point, or increase the water's temperature......
water
It would depend on the temperature of the water, or average kinetic energy. (KE) However, what you may be looking for is how much heat is needed to raise the KE, or temperature, of water. 4.184 kilojoules per gram is the heat required to raise the temperature of water 1 degree Celsius.
6480 calories8LBs X 15 gallons x 54 DEGREES = 6480 CALORIES
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.
The temperature of the water would be 4.58333 degrees Celsius higher.