How much heat must be added to 3kg of liquid water at 25 degrees C to change it into steam at 120 degrees C?
figure this in three parts:
heat to heat water from 25C to 100C
heat to vaporize water at 100C
heat to heat steam from 100C to 120C
add them
To calculate the amount of heat required to change liquid water into steam, we need to consider two stages: heating the water from 25°C to its boiling point, and then vaporizing it from liquid to steam.
The first stage is the heating phase, where we need to calculate the heat required to raise the temperature of the water from 25°C to its boiling point. This can be done using the specific heat capacity equation:
Q1 = m * c * ΔT
Here,
Q1 represents the heat required for heating the water,
m is the mass of the water (3kg in this case),
c is the specific heat capacity of water (approximately 4.186 J/g°C),
ΔT is the change in temperature (boiling point - initial temperature).
The boiling point of water is 100°C; therefore, ΔT for this stage is:
ΔT1 = 100°C - 25°C = 75°C
Plugging in the values, we get:
Q1 = 3kg * 4.186 J/g°C * 75°C
Q1 = 940.35 kJ
Now, we move on to the second stage, which involves the phase change from liquid to steam. The amount of heat required to achieve this is given by the formula:
Q2 = m * L
Here,
Q2 represents the heat required for vaporization,
m is the mass of the water (3kg in this case),
L is the specific heat of vaporization of water (approximately 2260 kJ/kg).
Plugging in the values, we get:
Q2 = 3kg * 2260 kJ/kg
Q2 = 6780 kJ
To find the total heat required, we sum up Q1 and Q2:
Total heat = Q1 + Q2
Total heat = 940.35 kJ + 6780 kJ
Total heat ≈ 7720.35 kJ
Therefore, approximately 7720.35 kJ of heat must be added to change 3kg of liquid water at 25°C into steam at 120°C.