200g of steam at 110°C is condensed, and the resulting water is frozen into ice at 0°C. How much heat was removed?
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To calculate the heat removed during the condensation of steam and subsequent freezing of water, we need to determine the amount of heat involved in each step.
First, we need to calculate the heat released during the condensation of steam. The heat released during condensation can be calculated using the formula:
Q1 = m * Latent heat of vaporization
where:
Q1 = Heat released during condensation (in joules)
m = Mass of steam (in kilograms)
Latent heat of vaporization = 2,260,000 J/kg (for water)
Converting the mass of steam from grams to kilograms:
m = 200 g = 200/1000 = 0.2 kg
Calculating the heat released during condensation:
Q1 = 0.2 kg * 2,260,000 J/kg = 452,000 J
Next, we need to calculate the heat removed during the freezing of water. The heat removed during freezing is given by the formula:
Q2 = m * Latent heat of fusion
where:
Q2 = Heat removed during freezing (in joules)
m = Mass of water (in kilograms)
Latent heat of fusion = 334,000 J/kg (for water)
Since the mass of water before freezing is the same as the mass of steam condensate, we can use m = 0.2 kg.
Calculating the heat removed during freezing:
Q2 = 0.2 kg * 334,000 J/kg = 66,800 J
Finally, to determine the total heat removed, we add the heat released during condensation and the heat removed during freezing:
Total heat removed = Q1 + Q2
Total heat removed = 452,000 J + 66,800 J
Total heat removed = 518,800 J
Therefore, approximately 518,800 joules of heat was removed during the condensation of steam at 110°C and subsequent freezing of water at 0°C.
To find out how much heat was removed, we need to consider the different stages of the process.
First, we need to determine the heat required to condense the steam at 110°C into water at 110°C. The heat required to condense the steam is given by the formula:
Q1 = mass * latent heat of vaporization
where Q1 is the heat required, and the mass is given as 200g. The latent heat of vaporization for water is approximately 2260 J/g. Plugging in the values, we get:
Q1 = 200g * 2260 J/g = 452,000 J
So, the heat required to condense the steam is 452,000 J.
Next, we need to determine the heat required to cool down the water from 110°C to 0°C. The heat required to cool down the water is given by the formula:
Q2 = mass * specific heat capacity * temperature change
where Q2 is the heat required, the mass is given as 200g, the specific heat capacity of water is approximately 4.18 J/g°C, and the temperature change is 110°C - 0°C = 110°C. Plugging in the values, we get:
Q2 = 200g * 4.18 J/g°C * 110°C = 92,360 J
So, the heat required to cool down the water is 92,360 J.
Lastly, we need to determine the heat required to freeze the water into ice at 0°C. The heat required to freeze the water is given by the formula:
Q3 = mass * latent heat of fusion
where Q3 is the heat required, and the mass is given as 200g. The latent heat of fusion for water is approximately 334 J/g. Plugging in the values, we get:
Q3 = 200g * 334 J/g = 66,800 J
So, the heat required to freeze the water into ice is 66,800 J.
Now, to find the total heat removed, we sum up all the heat values:
Total heat removed = Q1 + Q2 + Q3
= 452,000 J + 92,360 J + 66,800 J
= 611,160 J
Therefore, approximately 611,160 J of heat was removed during the process.
cool the steam to 100ºC
condense the steam
cool the water to 0ºC
freeze the water
two specific heats
heat of vaporization, and heat of fusion