How much heat is given up when 20 g of steam at 100 deg C is condensed and cooled to 20 deg.C?
Total Q= -160.5 cal
Well, it really depends on how good your steam is at breaking up with its heat. Some steam might give up just a little heat and remain friends with the temperature, while others might give up all the heat and end up in a cold, lonely place like my ex's heart.
But to answer your question, the amount of heat given up can be calculated using the equation:
Q = m * C * ΔT
Where:
Q is the heat given up,
m is the mass of the steam (20 g),
C is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (100°C - 20°C).
So, let's crunch some numbers and find out how much heartbreak we're dealing with!
To determine the amount of heat given up when steam is condensed and cooled, we need to consider two separate processes: the heat released during condensation and the heat released during cooling.
1. Heat released during condensation:
To convert steam to water, the steam needs to release its latent heat of vaporization. The latent heat of vaporization of water is approximately 2260 kJ/kg. However, we have been given the mass in grams, so we need to convert it to kilograms.
Given:
Mass of steam, m = 20 g = 0.02 kg
Latent heat of vaporization, Lv = 2260 kJ/kg
The heat released during condensation can be calculated using the formula:
Q = m * Lv
Q = 0.02 kg * 2260 kJ/kg = 45.2 kJ
Therefore, the heat released during condensation is 45.2 kJ.
2. Heat released during cooling:
To calculate the heat released during cooling, we need to consider the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 kJ/kg·°C.
Given:
Mass of water, m = 20 g = 0.02 kg
Initial temperature, Ti = 100°C
Final temperature, Tf = 20°C
Specific heat capacity, c = 4.18 kJ/kg·°C
The heat released during cooling can be calculated using the formula:
Q = m * c * ΔT
Where ΔT is the change in temperature: ΔT = Tf - Ti
ΔT = 20°C - 100°C = -80°C
Q = 0.02 kg * 4.18 kJ/kg·°C * (-80°C) = -66.56 kJ
Note that the negative sign indicates that heat is being released during cooling.
Therefore, the heat released during cooling is 66.56 kJ.
The total heat released when 20 g of steam at 100°C is condensed and cooled to 20°C is the sum of the heat released during condensation and cooling:
Total heat released = Heat released during condensation + Heat released during cooling
Total heat released = 45.2 kJ + (-66.56 kJ)
Total heat released = -21.36 kJ
The negative sign indicates that heat is being released during the process. Therefore, the amount of heat given up is approximately 21.36 kJ.
q1 = heat lost in condensing steam
q1 = mass steam x heat vaporization
q2 = heat lost in cooling from 100 C to 20 C.
q2 = mass water x specific heat water x (Tfinal-Tinitial)
Total Q = q1 + q2.