What is the energy released when steam at 25 g of steam at 10 degree celcius is cooled to ice at-10 degree celcius
work
heat change for the following changes calculated
150 degree Celsius steam --150 degree water
150 degree water to 0 degree water
100 degreewater-- 0 degree ice
0 degree ice----10 degree ice
but i am notgetting the correct answer
Instead of me guessing what you did why not show your work and let me find the error.
To calculate the energy released when steam at 25 g of steam at 10 degrees Celsius is cooled to ice at -10 degrees Celsius, we can follow these steps:
1. Determine the energy required to heat the steam from 10 degrees Celsius to its boiling point, which is 100 degrees Celsius. This can be done using the formula:
Q = m * c * ΔT
Where:
Q is the heat energy
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For steam, the specific heat capacity (c) is 2.03 J/g°C. Therefore, the energy required to heat the steam from 10 degrees Celsius to boiling is:
Q1 = 25 g * 2.03 J/g°C * (100°C - 10°C) = 38,225 J
2. Next, we need to calculate the energy released when the steam condenses to water at 100 degrees Celsius. This can be done using the formula:
Q2 = m * ΔHv
Where:
Q2 is the heat energy
m is the mass of the substance
ΔHv is the heat of vaporization of the substance
For water, the heat of vaporization (ΔHv) is 40.7 kJ/mol, which is equivalent to 40.7 J/g. Therefore, the energy released when the steam condenses to water at 100 degrees Celsius is:
Q2 = 25 g * 40.7 J/g = 1017.5 J
3. Calculate the energy required to cool the water from 100 degrees Celsius to 0 degrees Celsius using the formula:
Q3 = m * c * ΔT
For water, the specific heat capacity (c) is approximately 4.18 J/g°C. Therefore, the energy required to cool the water from 100 degrees Celsius to 0 degrees Celsius is:
Q3 = 25 g * 4.18 J/g°C * (0°C - 100°C) = -10450 J
Note that the negative sign indicates the release of energy.
4. Calculate the energy required to cool the water from 0 degrees Celsius to -10 degrees Celsius using the same formula as step 3:
Q4 = m * c * ΔT
For water, the specific heat capacity remains the same at 4.18 J/g°C. Therefore, the energy required to cool the water from 0 degrees Celsius to -10 degrees Celsius is:
Q4 = 25 g * 4.18 J/g°C * (-10°C - 0°C) = -1045 J
5. Add up all the energy changes to determine the total energy released:
Total energy released = Q1 + Q2 + Q3 + Q4
= 38225 J + 1017.5 J - 10450 J - 1045 J
= 26847.5 J
Therefore, the energy released when steam at 25 g of steam at 10 degrees Celsius is cooled to ice at -10 degrees Celsius is approximately 26847.5 J.