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The specific heat of water is 4.2 joules over g C if you wish to heat 300 grams of water from 30 C to 100 C how much heat will be required?
Wiki User
∙ 8y ago
To solve this you would use the equation q=mCs deltaT, in which q is the amount of energy needed for the reaction (in Joules), m is the mass of the substance in grams, Cs is the specific heat, and deltaT is the change in temperature of the reaction.
In this case,
q = what you are looking for
Cs = 4.184J/g*C
m = 46.0 grams
delta T = 100.0 degrees celsius
q = 46.0g x 4.184j/g*c x 100.0 *c
q = 19246.4 Joules
Wiki User
∙ 14y ago
Wiki User
∙ 13y ago4.184 Joules are required to heat up 1.000 gram of liquid water for 1.000 oC.
So for 46.0 grams and from 0.0 to 100.0 oC the answer is:
418.4(J/g.oC) * 46.0(g) * 100.0(oC) = 19246.4 J = 19.2 kJ(rounded to the least significant number -3-)
Wiki User
∙ 8y agoThe answer that you will get at the end is 88200 J.
Wiki User
∙ 13y ago67.6 degrees C
Wiki User
∙ 6y agoThe necessary heat is 229 joules.
Wiki User
∙ 15y ago69.6
Wiki User
∙ 12y agoa plus 0.88
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The temperature of 15 grams of water is increased by 3.0 Celsius degrees How much heat in Joules was absorbed by the water?
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
What is the total number of joules required to melt 100 grams of ice at 0 c to liquid water at 0 c?
226,ooo j
How much energy is required to raise the temperature of one liter of water 1 degree centigrade?
The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
How many joules of energy are needed to heat 25.0 grams of steam from 100.0 degrees Celsius to 29.25 degrees Celsius?
Use this formula. q(in Joules) = Mass * specific heat * change in temperature I will use specific heat of water at 25 C. You can look up specific heat of steam. You say " heat to " so I assume you have final and initial heat backwards. q = 25.0 grams H2O * 4.180 J/gC * (100.0 C - 29.25 C ) q = 7393 Joules
How many joules of energy are necessary to heat sample of water with a mass of 46.0 grams from 0.0 to 100 C?
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.
Specific heat of water equals?
4.18 joules over grams n temp
What is the total amount of heat energy in joules absorbed by 25.0 grams of water when the temperature of the water increased from 24.0 C to 36.0 C?
2520 Joules = (X)(4.180J/gC)(30C-10C) 2520J = 83.6X 30.14 grams of water.
The temperature of 15 grams of water is increased by 3.0 Celsius degrees How much heat in Joules was absorbed by the water?
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
How much heat is required to raise the temperature to 5.0 mL of water from 2.50 degrees Celsius to 75 degrees Celsius?
Density = grams/ml 1.00 g/ml = X g/5.0 ml = 5.0 grams water ============== q(joules) = mass * specific heat * change in temp. q = (5.0 grams)(4.180 J/gC)(75 C - 2.50 C) = 1515.25 Joules ---------------------------------/4.184 = 362 calories -------------------
What is the total number of joules required to melt 100 grams of ice at 0 C to liquid water at C?
226,ooo j
How many joules of energy are necessary to heat a smaple of water with a mass of 46.0 grams from 0.0 c to 100.0 c?
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
What is the total number of joules required to melt 100 grams of ice at 0 c to liquid water at 0 c?
226,ooo j
How much energy is required to raise the temperature of one liter of water 1 degree centigrade?
The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 celsius to 100.0?
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
Suppose you want to heat 40g of water by 20c how many joules of heat are required?
334.8 Joules
How many joules are needed to melt 2 grams of water?
2,26 Kj are necessary
How do you increase the the temperature of 500 grams of water from 20 degrees Celsius to 100 degrees Celsius?
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
