how many joules of energy are needed to change 275 g of water from 15 degrees C to steam at 100 degrees C
please tell me if this is set up right (275g)(4.184J/1 g Degree C)(85Degrees C)
To calculate the energy needed to change the temperature of water from 15°C to steam at 100°C, we need to consider two steps: heating the water from 15°C to 100°C, and then converting it from liquid to steam.
1. Heating the water:
The specific heat capacity of water is approximately 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.
So, to heat 275 grams of water from 15°C to 100°C, we can use the formula:
Q = m * c * ΔT
Where:
Q = energy in joules
m = mass of water in grams
c = specific heat capacity of water in J/g°C
ΔT = change in temperature in °C
Q = 275 g * 4.18 J/g°C * (100°C - 15°C)
Calculating this: Q = 275 * 4.18 * 85 = 97400.5 joules (or approximately 97400 joules)
2. Converting to steam:
The heat of vaporization of water is approximately 2260 J/g. This value represents the amount of energy required to convert 1 gram of water at its boiling point into steam.
To calculate the energy needed to convert 275 grams of water to steam, we can use the formula:
Q = m * Hv
Where:
Q = energy in joules
m = mass of water in grams
Hv = heat of vaporization in J/g
Q = 275 g * 2260 J/g
Calculating: Q = 621500 joules
Adding the two calculated values, the total energy required is:
97400 joules + 621500 joules = 718900 joules
Therefore, approximately 718900 joules of energy are needed to change 275 grams of water from 15°C to steam at 100°C.
To determine the amount of energy required, we need to calculate the heat energy needed to raise the temperature of water from 15 degrees Celsius to its boiling point and then the heat energy needed to change it from water to steam.
To calculate the heat energy required to raise the temperature of water:
1. Determine the mass of water (m): 275 g.
2. Find the specific heat capacity of water (c): It is approximately 4.18 J/g°C.
3. Calculate the temperature change (∆T): 100°C - 15°C = 85°C.
4. Use the formula: q = m * c * ∆T, where q is the heat energy.
q = 275 g * 4.18 J/g°C * 85°C.
The above calculation will give us the amount of heat energy needed to raise the temperature of water to its boiling point.
Now, to calculate the heat energy required to change water from liquid to steam:
1. Determine the heat of vaporization (Hvap) of water: It is approximately 2260 J/g.
2. Calculate the mass of water transformed to steam, which is the same as the initial mass of water: 275 g.
The heat energy (q) required to change water to steam is: q = m * Hvap = 275 g * 2260 J/g.
Finally, to get the total energy needed to change 275 g of water from 15 degrees C to steam at 100 degrees C, add the heat energy values obtained from both calculations:
Total energy = heat energy to raise the temperature + heat energy to change from water to steam.
Total energy = (275 g * 4.18 J/g°C * 85°C) + (275 g * 2260 J/g).
You have half of it right. The other half you have ignored.
q1 = heat needed to raise T of water from 15C to 100 C is what you have calculated.
q2 = heat needed to change T of water at 100 C to steam at 100 C.
That is q2 = mass H2O x heat vaporization.