42 J
Use this formula.
q(in Joules) = mass * specific heat of water * change in temp.(final - initial )
q = nCT
q = (46.0 grams)(4.180 J/gC)(100 C - 0 C)
= 19228 Joules
correct for significant figures
Q = mCΔt, where Q is energy added or lost in Joules, m is mass in grams, C is specific heat, Δtis change in temperature in oC.
m = 2.0g
Δt = 5.0 oC
CH2O = 4.186 Joules/gram oC
Q = 2.0g x 4.180J/g•oC x 5.0 oC = 42 Joules (rounded to two significant figures)
For each Celsius degree of heating, you must add 4.184 joules per gram,
or 192.4 joules for the 46 gram sample.
1 calories (or 4,184 joule) for 1 g and for each Celsius degree.
Q (heat) = mass*Cp*Temp. Change (where Cp is specific heat)
Q = 5.0 grams * 4.18 J/goC * (30-20oC)
=5*4.18*10 = 209 J
46 calories (or 192, 464 joules) for each Celsius degree.
q(joules) = mass * specific heat * change in temperature q = (500 grams H2O)(4.180 J/goC)(100o C - 20o C) = 1.7 X 105 joules ================add this much heat energy to the water
4.1858 joules of energy will raise the temperature of 1 g of water by 1oC. Thus, 4.1858 * 955 * 80 = 319795.12 joules of energy is required to raise the temperature of 955 g of water by 1oC.
10 degrees Celsius
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams
Heat is measured in unit of what...
The answer is 53,683 kJ.
15.37684 joules
15480.80
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
46 calories (or 192, 464 joules) for each Celsius degree.
46 calories (or 192, 464 joules) for each Celsius degree.
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.
3.8 x 10^5 Joules
I will use this formula. Some conversion will be required. ( I only know specific heat iron in J/gC ) q(Joules) = mass * specific heat * change in temperature Celsius 3 kilograms cast iron = 3000 grams q = (3000 g)(0.46 J/gC)(120 C - 30 C) = 124200 Joules (1 kilojoule/1000 joules) = 124.2 kilojoules of energy needed ===========================
Approx 2940 Joules.
Can you help