# An Experiment to Determine the Heating and Cooling Curves of Water

**Pages:**5 (1241 words)

**Published:**May 6, 2013

(pg 1 of 2)

Heating and Cooling Curves

What happens when we heat a sample of ice that is initially at -15°C? The addition of heat causes the temperature of the ice to increase. As long as the temperature is below 0°C, the sample remains frozen. When the temperature reaches 0°C (the melting point of water), the ice begins to melt. Because melting is an endothermic process, the heat we add at 0°C is used to convert ice to water and the temperature remains constant until all the ice has melted. Once we reach this point, any further addition of heat causes the temperature of the liquid water to increase. You observed a phase change process for ice (solid water) being melted, heated and then boiled in class. A graph of the temperature of the system versus the amount of heat added (or temp vs time if the heat is added at a constant rate) is called a heating curve. You made a heating curve for water as it melted, then heated then vaporized in LAD 5 and a cooling curve for paradichlorobenzene as it goes through a cooling-freezing-cooling process in LAD 6.

Understand each Segment on the Heating Curve below

AB Heating ice from -15°C to 0°C is represented by the sloping line segment AB Converting the ice at 0°C to water at 0°C is the horizontal segment BC. Additional heat increases the temperature of the water until the temperature reaches 100°C during segment CD. The heat is then used to convert water to steam at a constant temperature of 100°C during segment DE. The steam is then heated to its final temperature of 125°C in the sloping segment EF.

Heating Curve for Water

120 100 temp (ºC) 80 60 40 20 0

F D E

BC

CD

DE

B

C

EF

-20 A

0

5000

10000 15000 20000 25000 30000 35000

heat (Joules)

Energy Calculations for the Sloping Segments of the Heating (or Cooling) Curve We can calculate the energy change of the system for each of the segments of a heating curve. In segments AB, CD, and EF we are heating a single phase from one temperature to another. Recall equation 3.1 in NS 3, used to calculate the amount of heat needed to raise the temperature of a substance; the product of the specific heat capacity (SHC), mass, and temperature change. The greater the specific heat of a substance, the more heat we must add to accomplish a certain temperature increase. SHC * mass * ∆T = heat lost or gained segment AB: 2.1 J/gºC * 10 g * 15ºC = 315 J segment CD: 4.18 J/gºC * 10 g * 100ºC = 4,180 J segment EF: 1.7 J/gºC * 10 g * 25ºC = 425 J the H2O is ice so the SHC of ice must be used the H2O is liquid so the SHC of liquid water must be used the H2O is steam so the SHC of steam must be used

Because the specific heat of liquid water is greater than that of ice and steam, the slope of segment CD or EF is less than that of segment AB; we must add more heat to water to achieve a 1°C temperature change than is needed to warm the same quantity of ice or steam by 1°C.

NS 10.5 (pg 2 of 2 )

Heating and Cooling Curves

Energy Calculations for the Plateau Segments of the Heating (or Cooling) Curve ∆Hfusion and ∆Hvaporization - available on the bottom of the lemon yellow sheet In segments BC and DE we are converting one phase to another at a constant temperature. The temperature remains constant during these phase changes, because the added energy is used to overcome the attractive forces between molecules rather than to increase their average kinetic energy. For segment BC, in which ice is converted to water, the enthalpy change can be calculated by using ∆Hfus, while for segment DE we can use ∆Hvap Thus the equation for calculating the energy change during a phase change is given below: ∆Hvap or fus kJ/mole * * moles moles = = heat lost or gained kJ (kiloJoules)

be sure and notice that the units on the right side again cancel out to equal the units on the left side of the equation.

Be sure and take note that in the equation above there is no ∆T because during a phase change, the...

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