How much heat is added to 10.5 g of ice at -20.0°C to convert it to steam at 123.5°C?
do this in parts
heat to warm ice to 0C
heat to melt ice at 0C
heat to heat water from 0 to 100C
heat to vaporize water at 100C
heat to heat steam from 100Cto final temp
add the heats
To calculate the heat added to convert ice at -20.0°C to steam at 123.5°C, we need to consider the following steps:
1. Heat required to raise the temperature of ice from -20.0°C to the melting point (0°C).
2. Heat required to change the ice at 0°C to water at 0°C (latent heat of fusion).
3. Heat required to raise the temperature of water from 0°C to the boiling point at 100°C.
4. Heat required to change the water at 100°C to steam at 100°C (latent heat of vaporization).
5. Heat required to raise the temperature of steam from 100°C to 123.5°C.
Now, let's calculate each step:
Step 1: Heat required to raise the temperature of ice from -20.0°C to 0°C.
The specific heat capacity of ice is 2.09 J/g°C. The temperature change is 0°C - (-20.0°C) = 20.0°C.
Heat = mass * specific heat capacity * temperature change
Heat = 10.5 g * 2.09 J/g°C * 20.0°C = 439.8 J
Step 2: Heat required to change the ice at 0°C to water at 0°C (latent heat of fusion).
The latent heat of fusion for ice is 334 J/g.
Heat = mass * latent heat of fusion
Heat = 10.5 g * 334 J/g = 3507 J
Step 3: Heat required to raise the temperature of water from 0°C to 100°C.
The specific heat capacity of water is 4.18 J/g°C.
The temperature change is 100°C - 0°C = 100°C.
Heat = mass * specific heat capacity * temperature change
Heat = 10.5 g * 4.18 J/g°C * 100°C = 4389 J
Step 4: Heat required to change the water at 100°C to steam at 100°C (latent heat of vaporization).
The latent heat of vaporization for water is 2260 J/g.
Heat = mass * latent heat of vaporization
Heat = 10.5 g * 2260 J/g = 23730 J
Step 5: Heat required to raise the temperature of steam from 100°C to 123.5°C.
The specific heat capacity of steam is 2.03 J/g°C.
The temperature change is 123.5°C - 100°C = 23.5°C.
Heat = mass * specific heat capacity * temperature change
Heat = 10.5 g * 2.03 J/g°C * 23.5°C = 489.345 J
Finally, we can add up the heat from each step:
Total Heat = Heat from step 1 + Heat from step 2 + Heat from step 3 + Heat from step 4 + Heat from step 5
Total Heat = 439.8 J + 3507 J + 4389 J + 23730 J + 489.345 J
Total Heat = 33555.145 J
Therefore, the heat added to convert 10.5 g of ice at -20.0°C to steam at 123.5°C is approximately 33555.145 J.
To calculate the amount of heat added to convert ice at -20.0°C to steam at 123.5°C, you need to consider the following steps:
1. Calculate the heat required to raise the temperature of the ice from -20.0°C to 0°C.
2. Calculate the heat required to melt the ice at 0°C.
3. Calculate the heat required to raise the temperature of the water from 0°C to 100°C.
4. Calculate the heat required to vaporize the water at 100°C.
5. Calculate the heat required to raise the temperature of the steam from 100°C to 123.5°C.
Let's calculate each step individually:
1. Calculate the heat required to raise the temperature of the ice from -20.0°C to 0°C (using the specific heat capacity of ice):
Q1 = m * Cp * ΔT
m = 10.5 g (mass of ice)
Cp = 2.09 J/g°C (specific heat capacity of ice)
ΔT = (0°C - (-20.0°C)) = 20.0°C
Q1 = 10.5 g * 2.09 J/g°C * 20.0°C = 439.8 J
2. Calculate the heat required to melt the ice at 0°C (using the heat of fusion of ice):
Q2 = m * ΔHfus
ΔHfus = 334 J/g (heat of fusion of ice)
Q2 = 10.5 g * 334 J/g = 3510 J
3. Calculate the heat required to raise the temperature of the water from 0°C to 100°C (using the specific heat capacity of water):
Q3 = m * Cp * ΔT
m = 10.5 g (mass of water)
Cp = 4.18 J/g°C (specific heat capacity of water)
ΔT = (100°C - 0°C) = 100°C
Q3 = 10.5 g * 4.18 J/g°C * 100°C = 4395 J
4. Calculate the heat required to vaporize the water at 100°C (using the heat of vaporization of water):
Q4 = m * ΔHvap
ΔHvap = 2260 J/g (heat of vaporization of water)
Q4 = 10.5 g * 2260 J/g = 23730 J
5. Calculate the heat required to raise the temperature of the steam from 100°C to 123.5°C (using the specific heat capacity of steam):
Q5 = m * Cp * ΔT
m = 10.5 g (mass of steam)
Cp = 2.03 J/g°C (specific heat capacity of steam)
ΔT = (123.5°C - 100°C) = 23.5°C
Q5 = 10.5 g * 2.03 J/g°C * 23.5°C = 493 J
Finally, add up all the calculated values to get the total heat added:
Total heat added = Q1 + Q2 + Q3 + Q4 + Q5
Total heat added = 439.8 J + 3510 J + 4395 J + 23730 J + 493 J
Total heat added ≈ 33068 J
Therefore, approximately 33,068 Joules of heat is added to convert 10.5 grams of ice at -20.0°C to steam at 123.5°C.