how much heat energy is released from cooling 35 grams of steam at 135 degrees celcius down to freezing it to negative 45 degrees celcius?
Note the correct spelling of celsius.
There are two equations you need to do this.
Within a phase use
mass x specific heat x (Tfinal-Tinitial)
For example to change T of water from liquid water at 100 C to liquid water at 0 C, it is
mass H2O x 4.184 x (0-100) = ?
At a phase change use
mass x heat fusion at freezing point
mass x heat vaporization at boiling point.
For example, to convert steam at 100 C to liquid at 100 C it will be
mass steam x heat vaporization.
Then add each part to find the total heat transferred.
To determine the amount of heat energy released when cooling steam from 135°C to -45°C, we need to calculate the change in heat energy using the specific heat capacity and the phase change of water.
First, we need to break this problem down into two parts:
1. Cooling the steam from 135°C to 0°C.
2. Freezing the water from 0°C to -45°C.
Let's start with the first part:
1. Cooling the steam from 135°C to 0°C:
The specific heat capacity of steam is approximately 2.03 J/g°C.
To find the energy released, we need to calculate the change in temperature:
ΔT = (Tfinal - Tinitial)
ΔT = (0°C - 135°C) = -135°C
Now we can calculate the energy (Q1) released using the specific heat capacity (c1) and the mass (m1) of steam:
Q1 = m1 * c1 * ΔT
Q1 = 35 g * 2.03 J/g°C * -135°C
Q1 = -9740.25 J (Note: The negative sign indicates heat release.)
Next, moving on to the second part:
2. Freezing the water from 0°C to -45°C:
The specific heat capacity of water is approximately 4.18 J/g°C.
Again, we need to calculate the change in temperature:
ΔT = (Tfinal - Tinitial)
ΔT = (-45°C - 0°C) = -45°C
Now we can calculate the energy (Q2) released using the specific heat capacity (c2) and the mass (m2) of water:
Q2 = m2 * c2 * ΔT
Q2 = 35 g * 4.18 J/g°C * -45°C
Q2 = -2654.25 J (Note: The negative sign indicates heat release.)
Finally, to find the total heat energy released, we sum up Q1 and Q2:
Total heat energy released = Q1 + Q2
Total heat energy released = -9740.25 J + -2654.25 J
Total heat energy released ≈ -12394.5 J
Therefore, approximately 12,394.5 Joules (J) of heat energy is released when cooling 35 grams of steam from 135°C to freezing it to -45°C.