How many litres of water at 100°C are vaporized by the addition of 5.00 MJ of heat
5 MJ = 5E6 J
q = mass H2O in grams x delta H vap in J/g.
5E6J = mass H2O x delta H vap.
Look up delta H vaporization/condensation, substitute and solve for mass H2O, then look up the density of H2O at 100 C and convert grams H2O to mL H2O, then convert to L.
Would I need to multiply this to get the mass?
(2.26*10^6)(5*10^6)
I don't know what you're asking?
5E6 J = grams H2O x 2260 J/g
grams H2O = 5E6J/2260 J/g = ? grams H2O
Then use density H2O at 100 C to convert to L H2O.
So the mass H20 would be
5E6/2260 X 2260 = 5E6 grams?
water density is 0.9584 L from 100°C
I mean 958.4 kg/m3 from 100°C
No. mass H2O = 5E6/2260 =?g just as I said above.
Then mass = volume x density. Substitute ?g for mass and the density and solve for volume.
5E6/2260=2.21g then converted 2.21g to 0.00221L
0.00221X958.4= 2.119064
To find the number of liters of water that are vaporized by the addition of 5.00 MJ of heat, we need to understand the concept of specific heat and heat of vaporization.
1. Start by determining the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g·°C. This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
2. Convert the given heat energy of 5.00 MJ to Joules. Since 1 MJ (megajoule) is equal to 1,000,000 Joules, 5.00 MJ is equal to 5,000,000 Joules.
3. Next, calculate the amount of heat required to raise the temperature of the water from 100°C to its boiling point. The boiling point of water is 100°C. Since the specific heat capacity of water is given in grams (per gram), we need to convert liters to grams. The density of water is approximately 1 g/mL, so 1 liter of water is equal to 1000 grams. Multiply the specific heat capacity (4.18 J/g·°C) by the mass of the water to get the amount of heat required.
Q1 = (4.18 J/g·°C) * (1000 g) * (100°C - 0°C) = 418,000 J
4. Now, calculate the amount of heat required for the phase change from liquid to vapor. The heat of vaporization of water is 2.26 MJ/kg, which means it takes 2.26 MJ of energy to vaporize 1 kilogram of water.
Q2 = 2.26 MJ
5. Convert the heat of vaporization to Joules:
Q2 = 2.26 MJ * 1,000,000 J/MJ = 2,260,000 J
6. Finally, calculate the total heat required to vaporize the water. Add Q1 and Q2 together:
Total heat required = Q1 + Q2 = 418,000 J + 2,260,000 J = 2,678,000 J
Since we are given 5.00 MJ (5,000,000 J) of heat, we need to subtract the heat required for raising the temperature from 0°C to 100°C (Q1) from the total heat required (2,678,000 J) to find the heat used for vaporization.
Heat used for vaporization = Total heat required - Q1 = 2,678,000 J - 418,000 J = 2,260,000 J
7. Divide the heat used for vaporization by the heat of vaporization per kilogram (2,260,000 J) to find the mass of water vaporized:
Mass = Heat used for vaporization / Heat of vaporization per kilogram
Mass = 2,260,000 J / 2,260,000 J = 1 kilogram
Since 1 liter of water weighs 1 kilogram, the answer is 1 liter. Therefore, 1 liter of water at 100°C will be vaporized by the addition of 5.00 MJ of heat.