Compute the amount of heat is needed to raise the temperature of 1,200 grams of ice at -30 degrees C to 1,200 grams of water vapor at 140 degrees C. Show work.
To compute the amount of heat needed to raise the temperature, we will need to consider the following steps:
Step 1: Calculate the heat required to raise the temperature of ice to its melting point.
Step 2: Calculate the heat required to melt the ice.
Step 3: Calculate the heat required to raise the temperature of water from its melting point to its boiling point.
Step 4: Calculate the heat required to vaporize the water.
Step 5: Calculate the heat required to raise the temperature of steam to the final temperature.
Step 1:
The specific heat capacity of ice is 2.09 J/g°C.
The temperature change is given by:
Temperature change = final temperature - initial temperature
Temperature change = 0°C - (-30°C) = 30°C
Heat required = mass of ice × specific heat capacity × temperature change
Heat required = 1,200 g × 2.09 J/g°C × 30°C
Heat required = 74,760 J
Step 2:
The heat of fusion (latent heat) for water is 333.55 J/g.
The heat required to melt the ice is given by:
Heat required = mass of ice × heat of fusion
Heat required = 1,200 g × 333.55 J/g
Heat required = 400,260 J
Step 3:
The specific heat capacity of water is 4.18 J/g°C.
Temperature change = melting point of water - 0°C = 0°C (since the ice is already at the melting point)
Heat required = mass of water × specific heat capacity × temperature change
Since the mass of water is the same as the mass of ice (1,200 g), the heat required is:
Heat required = 1,200 g × 4.18 J/g°C × 0°C
Heat required = 0 J
Step 4:
The heat of vaporization for water is 2260 J/g.
Heat required = mass of water × heat of vaporization
Heat required = 1,200 g × 2260 J/g
Heat required = 2,712,000 J
Step 5:
The specific heat capacity of steam/vapor is 2.03 J/g°C.
The temperature change is given by:
Temperature change = final temperature - boiling point of water
Temperature change = 140°C - 100°C = 40°C
Heat required = mass of vapor × specific heat capacity × temperature change
Heat required = 1,200 g × 2.03 J/g°C × 40°C
Heat required = 97,440 J
Now, we can calculate the total heat required by summing up the results from each step:
Total heat required = Heat required in step 1 + Heat required in step 2 + Heat required in step 3 + Heat required in step 4 + Heat required in step 5
Total heat required = 74,760 J + 400,260 J + 0 J + 2,712,000 J + 97,440 J
Total heat required = 3,284,460 J
Therefore, the amount of heat needed to raise the temperature of 1,200 grams of ice at -30 degrees C to 1,200 grams of water vapor at 140 degrees C is 3,284,460 J.
To compute the amount of heat needed to raise the temperature of a substance, we can use the formula:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in joules per gram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, we need to calculate the heat energy required to raise the temperature of 1,200 grams of ice at -30 degrees Celsius to 1,200 grams of water vapor at 140 degrees Celsius.
Step 1: Calculate the heat energy required to raise the temperature of the ice from -30 degrees Celsius to 0 degrees Celsius.
Q1 = mcΔT1
Given:
m = 1,200 g (mass of ice)
c (specific heat capacity of ice) = 2.09 J/g°C (This is the specific heat capacity of ice at temperatures close to its melting point.)
ΔT1 = 0°C - (-30°C) = 30°C
Q1 = (1,200 g) * (2.09 J/g°C) * (30°C)
Q1 = 74,520 J
Step 2: Calculate the heat energy required to melt the ice into water at 0 degrees Celsius.
Q2 = mL
Given:
m = 1,200 g (mass of ice)
L = 334 J/g (latent heat of fusion for ice)
Q2 = (1,200 g) * (334 J/g)
Q2 = 400,800 J
Step 3: Calculate the heat energy required to raise the temperature of water at 0 degrees Celsius to water vapor at 100 degrees Celsius.
Q3 = mcΔT3
Given:
m = 1,200 g (mass of water)
c (specific heat capacity of water) = 4.18 J/g°C
ΔT3 = 100°C - 0°C = 100°C
Q3 = (1,200 g) * (4.18 J/g°C) * (100°C)
Q3 = 502,560 J
Step 4: Calculate the heat energy required to convert water at 100 degrees Celsius to water vapor at 100 degrees Celsius.
Q4 = mL
Given:
m = 1,200 g (mass of water)
L = 2,260 J/g (latent heat of vaporization for water)
Q4 = (1,200 g) * (2,260 J/g)
Q4 = 2,712,000 J
Step 5: Calculate the heat energy required to raise the temperature of water vapor at 100 degrees Celsius to water vapor at 140 degrees Celsius.
Q5 = mcΔT5
Given:
m = 1,200 g (mass of water vapor)
c (specific heat capacity of water vapor) = 1.996 J/g°C
ΔT5 = 140°C - 100°C = 40°C
Q5 = (1,200 g) * (1.996 J/g°C) * (40°C)
Q5 = 95,808 J
Step 6: Sum up all the heat energies calculated in the previous steps.
Total heat energy required = Q1 + Q2 + Q3 + Q4 + Q5
Total heat energy required = 74,520 J + 400,800 J + 502,560 J + 2,712,000 J + 95,808 J
Total heat energy required = 3,785,688 J
Therefore, the amount of heat needed to raise the temperature of 1,200 grams of ice at -30 degrees Celsius to 1,200 grams of water vapor at 140 degrees Celsius is 3,785,688 joules (J).