calculate the amount of heat energy in joules to condense 125g of steam at 100 degrees celcius and cool to the liquid to 15 degrees celcius?
To calculate the amount of heat energy required to condense and cool the steam, we need to consider two steps:
Step 1: Condensing the steam
To condense the steam, we need to calculate the heat energy required using the formula:
Q = m * Hv
where:
Q is the heat energy (in joules)
m is the mass of the steam (in grams)
Hv is the heat of vaporization (also known as latent heat) of water, which is 40.7 kJ/mol or 2.26 kJ/g.
First, we convert the mass of steam from grams to kilograms:
m = 125 g = 0.125 kg
Now, we can calculate the heat energy required to condense the steam:
Q1 = m * Hv = 0.125 kg * 2.26 kJ/g = 0.2825 kJ
To convert kilojoules to joules, we multiply by 1000:
Q1 = 0.2825 kJ * 1000 = 282.5 J
Step 2: Cooling the liquid water
To calculate the heat energy required to cool the liquid water, we use the formula:
Q = m * Cp * ΔT
where:
Q is the heat energy (in joules)
m is the mass of the water (in grams)
Cp is the specific heat capacity of water, which is approximately 4.18 J/g°C
ΔT is the change in temperature (in °C)
First, we need to calculate the change in temperature:
ΔT = Tf - Ti
= 15°C - 100°C
= -85°C
Note: The negative sign indicates a temperature decrease.
Now, we can calculate the heat energy required to cool the water:
Q2 = m * Cp * ΔT
= 125 g * 4.18 J/g°C * -85°C
= -56125 J
Since we are calculating the amount of energy required to cool the water, we should take Q2 as a positive value:
Q2 = -(-56125 J) = 56125 J
Finally, we can calculate the total heat energy required:
Total Heat Energy = Q1 + Q2
= 282.5 J + 56125 J
= 56307.5 J
Therefore, the amount of heat energy needed to condense 125g of steam at 100°C and cool it to 15°C is approximately 56307.5 Joules.
To calculate the amount of heat energy required to condense steam and cool it to a lower temperature, you need to consider two separate processes: (1) condensing steam from 100°C to water at 100°C, and (2) cooling the water from 100°C to 15°C.
First, let's calculate the heat energy required to condense the steam at 100°C to water at 100°C. This process involves the heat of vaporization:
1. Find the heat energy required to convert 125g of steam to water at 100°C:
- The heat of vaporization for water is approximately 2260 J/g.
- Multiply the heat of vaporization by the mass of steam:
2260 J/g * 125g = 282,500 J (rounded to three significant figures)
Next, let's calculate the heat energy required to cool the water from 100°C to 15°C. This process involves the specific heat capacity:
2. Find the heat energy required to cool 125g of water from 100°C to 15°C:
- The specific heat capacity of water is approximately 4.18 J/g°C.
- Subtract the initial temperature from the final temperature to find the temperature difference:
100°C - 15°C = 85°C
- Multiply the specific heat capacity by the mass of water and the temperature difference:
4.18 J/g°C * 125g * 85°C = 44,818.75 J (rounded to three significant figures)
To find the total heat energy required, add the heat energy from both processes:
282,500 J + 44,818.75 J = 327,318.75 J
Therefore, the amount of heat energy required to condense 125g of steam at 100°C and cool it to 15°C is approximately 327,319 J (rounded to three significant figures).
Note the correct spelling of celsius.
q1 = heat released to condense steam at 100 C to liquid at 100 C.
q1 = mass x heat vaporization
q2 = heat released on cooling from 100 C to 15 C.
q2 = mass water s specific heat water x (Tfinal-Tinitial).
Total Q = q1 + q2.