Heating Curves

We can combine what we know about heat capacities of substances (solid, liquid, gas) and combine that with heats/enthalpies of transition (ΔHs) and make a heating curve for any substance. The most common heating curve substance is water. Below is a heating curve for water that shows all FIVE transition regions or zones.



Those values for water are the following:

Cs,ice = 2.09 J/g °C

Cs,water = 4.184 J/g °C

Cs,steam = 2.03 J/g °C

ΔHfusion = 334 J/g

ΔHvaporization = 2260 J/g

Traversing the Curve

A full blown "heating curve" question will generally cover more that one region on the curve above. So what you have to realize is that you will break the problem into the regions as described and calculate the heats (q's) for each region. You then sum up the heats for each of the regions in the problem. More complicated problems will envolve the possibility of two samples starting at two different parts of the curve, but ultimately reaching the same exact final point (temperature and composition). Examples follow.

Multi-region Heating Curve Problem

A sample of 5 cubes of ice from my freezer melted and warmed until the water is at 104 °F. My freezer compartment is at 0 °F and the cubes of ice are 25 grams each. Calculate the amount of heat to do this.


solution...

Step 1: Change temperture units to Celcius.

°C = (°F - 32)/1.8 which means that 0 °F = -17 °C and 104 °F = 40 °C

Step 2: Map out the regions and temperatures on the heating curve. This is your "map" for the problem.

Note that there are 3 regions to calculate heat. They are
1) heating the -17 °C ice to 0 °C ice
2) melting the 0 °C ice to 0 °C water
3) heating the 0 °C water to 40 °C water

Step 3: do the calculations for each step

q1 = m Cs,ice ΔTice = (125 g ice)(2.09 J/g°C)(17°C) = 4441 J

q2 = m ΔHfusion = (125 g ice)(334 J/g) = 41750 J

q3 = m Cs,water ΔTwater = (125 g ice)(4.184 J/g°C)(40°C) = 20920 J

Step 4: Sum the steps and convert to kJ

4441 + 41750 + 20920 = 67111 J =   67.1 kJ  



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