David K. answered • 03/24/22
Expert, Friendly Physics Tutor with 5000+ Hours Tutoring Experience
Hi Max -
There are a couple aspects to this problem - it's important to remember that heat needs to be removed to decrease the temperature of a substance, but more heat also needs to be removed to cause phase changes to occur. The heat extraction here occurs in several phases: first we remove enough heat to reduce the temperature of the steam to 100ºC, and then we continue removing heat to cause the 100°C steam (gaseous) to condense into 100°C liquid water (liquid). We then continue removing heat to bring the temperature of the liquid water down to 0°C, and then lastly, we have to remove a little more heat in order to cause the 0°C liquid water (liquid) to freeze into 0°C ice (solid) to finish the problem.
To calculate the amount of heat necessary to cause a particular amount of a substance to change in temperature by a particular amount, we use the equation
q = m * c * Δt
where q is the amount of heat removed in joules, m is the mass of your substance in grams (1500 grams in this case), c is the specific heat of the substance whose temperature needs to be changed (for the steam we are dealing with at this moment, it's 2.03 J/g°C, can be looked up online), and Δt is the change in temperature in Celsius.
For the first removal of heat that cools the steam down to 100°C, it looks like this (notice how the multiplication of units leaves you with a quantity in joules):
q = m * c * Δt
q = 1500g * 2.03 J/g°C * 10°C
q = 30450 J
To calculate how much heat we need to remove to condense the 100°C steam into 100°C liquid water, we need to know the heat of vaporization of water - the amount of heat that needs to be added to 1 gram of liquid water to vaporize it into steam, or equivalently (and more important for this problem), the amount of heat that needs to be removed from 1 gram of steam to condense it into liquid water. This value is 2260 J/g (look it up online), so the calculation looks like this:
q = heat of vaporization * mass of steam to be condensed
q = 2260 J/g * 1500g
q = 3390000 J
Now we have 100°C liquid water, which needs to be cooled to 0°C liquid water before it can be turned to ice. We use the same equation for a temperature change as above, except this time we use the specific heat value for liquid water rather than ice, which is 4.184 J/g°C (again, look it up if you need):
q = m * c * Δt
q = 1500g * 4.184 J/g°C * 100°C
q = 627600 J
Finally, we have 0°C liquid water which needs to be frozen into 0°C ice to complete the transformation. Much like we used the heat of vaporization above to calculate the amount of heat we have to remove to turn steam into liquid water, now we need to use a quantity called the heat of fusion, which tells the amount of heat we have to remove from 1 gram of liquid water to turn it into ice. The heat of fusion of liquid water is 334 J/g (look it up), and so our calculation looks like this:
q = heat of fusion * mass of liquid water to be frozen
q = 334 J/g * 1500g
q = 501000 J
Now that we know the amount of heat removed at each stage of the process, we just add them up to calculate the total amount of heat removed for the whole thing:
q = 30450 J + 3390000 J + 627600 J + 501000 J
q = 4549050 J
I hope this was helpful! Send me a message if you have any questions and I'll be happy to explain more.
