how many kilocalories of heat are released when 85g steam at 100 degrees C converted to ice at 0.
Three steps are involved: steam changing to water at 100 C; water at 100 C cooling to 0 C; water at 0 C freezing to ice at 0 C.
Steam changing to water:
Heat of vaporization of water is 539.4 cal/g
Heat = Heat of Vap. x mass = 539.4 cal/g x 75.0 g = 40,500 cal
Cooling water from 100 C to 0 C:
Heat = mass x specific heat x T change
Heat = 75.0 g x 1.00 cal/gC x 100 C = 7500 cal
Freezing of water at 0 C:
The heat of fusion of water is 79.7 cal/g
Heat = heat of fusion x mass = 79.7 cal/g x 75.0 g = 5980 cal
Total heat = 40,500 + 7500 + 5980 = 53980 cal or 5.40 x 10^4 cal (3 sig figs) = 54.0 kcal
The work above is correct except the process is a bit confusing and hard to follow. I teach this process to my Honors Chemistry students in a similar way, just try to start with your grams and use factor labeling for the easiest way to solve.
To calculate the amount of heat released when steam at 100 degrees Celsius is converted into ice at 0 degrees Celsius, you need to consider the heat required for each step of the phase change.
The specific heat capacity of water is 4.18 J/g°C, and the heat of vaporization for water is 2260 J/g.
First, let's calculate the heat required to cool down the steam from 100°C to 0°C:
Q1 = m * c * ΔT
Q1 = 85g * 4.18 J/g°C * (0°C - 100°C)
Q1 = 85g * 4.18 J/g°C * (-100°C)
Q1 = -35,470 J
The negative sign indicates that heat is being released during this process.
Next, let's calculate the heat released during the phase change from steam to water at 100°C:
Q2 = m * ΔHvap
Q2 = 85g * 2260 J/g
Q2 = 191,500 J
Finally, let's calculate the heat released to further cool down the water from 0°C to ice at 0°C:
Q3 = m * c * ΔT
Q3 = 85g * 4.18 J/g°C * (0°C - 0°C)
Q3 = 0 J
Since the temperature doesn't change during this process, no heat is exchanged.
The total heat released is the sum of Q1, Q2, and Q3:
Q_total = Q1 + Q2 + Q3
Q_total = -35,470 J + 191,500 J + 0 J
Q_total = 156,030 J
To convert from joules to kilocalories, 1 kilocalorie (kcal) is equal to 4184 joules:
Q_total_kcal = Q_total / 4184
Q_total_kcal = 156,030 J / 4184
Q_total_kcal ≈ 37.29 kcal
Therefore, approximately 37.29 kilocalories of heat are released when 85g of steam at 100 degrees Celsius is converted to ice at 0 degrees Celsius.
To find the number of kilocalories of heat released during the conversion of steam to ice, we need to consider two steps:
1. Cooling the steam from 100 degrees Celsius to 0 degrees Celsius.
2. Freezing the water at 0 degrees Celsius into ice.
Let's calculate the heat released in each step:
Step 1: Cooling the steam to 0 degrees Celsius.
To find the heat released during this step, we can use the specific heat capacity of water, which is approximately 4.186 J/g°C. However, since we want the answer in kilocalories, we need to convert the units.
1 kilocalorie (kcal) = 1000 calories (cal)
1 calorie (cal) = 4.186 joules (J)
So, 1 kilocalorie = 1000 * 4.186 joules = 4186 joules
Now, let's calculate the heat released by cooling the steam:
Q = m * c * ΔT
where:
Q = heat released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
m = 85 g (mass of steam)
c = 4.186 J/g°C (specific heat capacity of water)
ΔT = 100°C - 0°C = 100°C
Q = 85 g * 4.186 J/g°C * 100°C = 35,548 joules
To convert this to kilocalories:
Q (in kcal) = Q (in joules) / 4186 = 35,548 J / 4186 = 8.50 kcal (rounded to two decimal places)
Therefore, cooling the steam from 100°C to 0°C releases approximately 8.50 kilocalories of heat.
Step 2: Freezing the water at 0 degrees Celsius into ice.
The heat released during the phase change from water to ice is known as the heat of fusion. For water, the heat of fusion is approximately 334 J/g.
Now, let's calculate the heat released by freezing the water:
Q = m * ΔHf
where:
Q = heat released (in joules)
m = mass of the substance (in grams)
ΔHf = heat of fusion (in J/g)
m = 85 g (mass of steam)
ΔHf = 334 J/g (heat of fusion for water)
Q = 85 g * 334 J/g = 28,390 joules
To convert this to kilocalories:
Q (in kcal) = Q (in joules) / 4186 = 28,390 J / 4186 = 6.79 kcal (rounded to two decimal places)
Therefore, freezing the water into ice releases approximately 6.79 kilocalories of heat.
To find the total heat released during the conversion of steam to ice, we need to add the heat released in both steps:
Total heat released = Heat released in step 1 + Heat released in step 2
Total heat released = 8.50 kcal + 6.79 kcal = 15.29 kilocalories (rounded to two decimal places)
Therefore, approximately 15.29 kilocalories of heat are released when 85 grams of steam at 100 degrees Celsius is converted to ice at 0 degrees Celsius.