calculate the heat required to change 5 lbs of 32°f ice to 212f° steam
Change 5 lbs ice to grams. 5 lbs x (459.6 g/1 lb) = 2298 grams.
Change 32 F to C. That's 0o C.
Cjamge 212 F tp C/ Tjat's 100c C.
q1 = heat to melt the ice at zero C to liquid @ zero C.
q1 = mass ice x heat fusion = ? You will need to look heat fusion.
q2 = heat to move liquid water from zero C to 100 C
q2 = mass water x specific heat H2O x (Tfinal-Tinitial) = ? You will need to look up the specific heat H2O
q3 = heat to convert liquid H2O @ 100 C to steam @ 100 C.
q3 = mass H2O x heat vaporization = ? You will need to look up the heat of vaporization for water.
Total heat = Qtotal = q1 + q2 + q3 = ?
Post your work if you get stuck. When looking up the values above make sure to use units consistent with the other parts of the problem. For example, if you're using grams you want to use specific heat water in J/g*C
Well, well, well! Looks like we've got a classic "from ice to steam" situation here. Let's do some calculations and heat things up!
To solve this, we need to consider three stages: heating the ice to its melting point, melting the ice into water, and heating the resulting water to its boiling point and changing it into steam.
First, let's warm up that ice. The specific heat capacity of ice is about 0.5 calories/g°C. Since we have 5 lbs of ice (which is around 2268 grams), and we need to raise it from 32°F (or 0°C) to 0°C, we can calculate the heat required using the formula:
Q = m * c * ΔT
Q = 2268 g * 0.5 cal/g°C * 0°C = 0 calories (No heat required!)
Next, we need to melt the ice into water. The heat of fusion for water is 79.7 calories/g. So, the heat required to melt the ice can be calculated as:
Q = m * ΔHf
Q = 2268 g * 79.7 cal/g = 180,603.6 calories
Finally, let's bring that water to a boil and transform it into steam. The specific heat capacity of water is about 1 cal/g°C, and the heat of vaporization is approximately 540 calories/g. We need to heat the 2268 g of water from 0°C to 100°C and then convert it to steam at 100°C. Let's calculate:
Q = m * c * ΔT + m * ΔHv
Q = 2268 g * 1 cal/g°C * 100°C + 2268 g * 540 cal/g = 256,680 calories
Now, let's add up the results from each stage:
Total Heat Required = 0 calories (ice heating) + 180,603.6 calories (melting ice) + 256,680 calories (boiling water)
Total Heat Required = 437,283.6 calories
So, approximately 437,284 calories of heat would be required to transform 5 lbs of 32°F ice to 212°F steam. That's a lot of hot air, isn't it?
To calculate the heat required to change ice to steam, we need to consider the different phases of water and the corresponding heat required for each phase change.
Step 1: Heating the ice to its melting point
The first step is to raise the temperature of the ice from 32°F to its melting point. The specific heat capacity of ice is 0.5 cal/g·°C or 1 cal/g·°F.
First, let's convert the weight from pounds to grams:
5 lbs = 5 x 453.59 g = 2267.95 g
Next, calculate the heat required to raise the temperature of ice to its melting point using the formula:
Q1 = m * c * ΔT
where:
Q1 is the heat required (in calories)
m is the mass of the ice (in grams)
c is the specific heat capacity of ice (in cal/g·°C or cal/g·°F)
ΔT is the change in temperature (in °C or °F)
Q1 = 2267.95 g * 1 cal/g·°F * (0°F - 32°F)
Q1 = -72,572.8 calories
Step 2: Melting the ice
The heat required to melt 1 gram of ice at its melting point is called the heat of fusion, which is 79.7 cal/g.
Now, calculate the heat required to melt the ice using the formula:
Q2 = m * Hf
where:
Q2 is the heat required (in calories)
m is the mass of the ice (in grams)
Hf is the heat of fusion (in cal/g)
Q2 = 2267.95 g * 79.7 cal/g
Q2 = 180,186.22 calories
Step 3: Heating the water to its boiling point
The next step is to raise the temperature of the water from its melting point to its boiling point. The specific heat capacity of water is 1 cal/g·°C or 1 cal/g·°F.
Calculate the heat required to raise the temperature of water to its boiling point using the formula:
Q3 = m * c * ΔT
where:
Q3 is the heat required (in calories)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in cal/g·°C or cal/g·°F)
ΔT is the change in temperature (in °C or °F)
Q3 = 2267.95 g * 1 cal/g·°F * (212°F - 32°F)
Q3 = 361,918.4 calories
Step 4: Vaporizing the water
The heat required to vaporize 1 gram of water at its boiling point is called the heat of vaporization, which is 540 cal/g.
Now, calculate the heat required to vaporize the water using the formula:
Q4 = m * Hv
where:
Q4 is the heat required (in calories)
m is the mass of the water (in grams)
Hv is the heat of vaporization (in cal/g)
Q4 = 2267.95 g * 540 cal/g
Q4 = 1,223,815.3 calories
Step 5: Total heat required
To calculate the total heat required, add up the heats obtained in each step:
Total heat = Q1 + Q2 + Q3 + Q4
Total heat = -72,572.8 + 180,186.22 + 361,918.4 + 1,223,815.3
Total heat ≈ 1,693,346.12 calories
Therefore, approximately 1,693,346.12 calories of heat are required to change 5 lbs of 32°F ice to 212°F steam.
To calculate the heat required to change ice to steam, we need to consider the heat absorbed during each phase transition.
First, we need to determine the heat required to raise the temperature of ice from 32°F to its melting point (32°F to 32°F). The specific heat capacity of ice is 0.5 cal/g·°C. We know the mass of the ice is 5 lbs, which is approximately 2268 grams.
Using the formula for heat, Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate the heat required to raise the temperature of the ice:
Q_ice = m * c * ΔT = 2268 g * 0.5 cal/g·°C * (0°C - (-32°C))
Next, we need to consider the heat required to melt the ice at its melting point (32°F to 32°F water). The heat of fusion, which is the heat required to convert a substance from solid to liquid without a temperature change, for ice is 79.7 cal/g.
Using the formula Q = m * ΔH_fus, where ΔH_fus is the heat of fusion, we can calculate the heat required to melt the ice:
Q_fusion = m * ΔH_fus = 2268 g * 79.7 cal/g
Finally, we need to determine the heat required to raise the temperature of the water from its melting point to its boiling point (32°F to 212°F). The specific heat capacity of water is 1 cal/g·°C.
Calculating the heat required to raise the temperature of the water:
Q_water = m * c * ΔT = 2268 g * 1 cal/g·°C * (212°C - 0°C)
To summarize, the total heat required is:
Total Heat = Q_ice + Q_fusion + Q_water
You can substitute the values into the respective formulas and calculate the total heat in calories. Just ensure that the units of mass and temperature are consistent.