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*Lesson 7.4:  Why does the ocean hold heat so well?

  1. What did we learn last class about sea level rise? Complete the IMT for Lesson 3.graph|400

  2. Using the phase change graph for water above, why does the temperature stay constant during each phase change? 

as the phase changes, the energy that is part of that phase is lost as particles transition from one state to another. This means that liquid water cant go above 100 degrees because it would lose energy as it evaporated.

3.  It is estimated that the Greenland ice sheet has a mass of 2.9 million gigatons. How much energy would it take to melt it if the starting temperature is -29 oC? (1 Gigaton = 1x10^12kg). 

\(2.9*10^{6} gT= 2.9*10^{12}kg = 2.9*10^{21}g\)

2090 joules per kg C

\(q = mc\Delta T\)

\(2.9*10^{12} * 2090 * 29 = 1.7*10^{23}J\)

344 joules per gram of ice to melt to water

\(q = \Delta H*m\)

\(9.7*10^{23}\)

\(1.7*10^{23} + 9.7*10^{23} = 1.14 * 10^{24}\)

Demonstration: Heating Balloons

  1. Place a balloon above a flame. 

  2. What will happen when we place a balloon filled with air above a flame? 

  3. What will happen when we place a balloon filled with water above a flame? 

  4. Which is able to hold more heat - air or water? Support your answer based on your observations.

Investigation 1: Mixing Water at Different Temperatures and Volumes

MATERIALS:

  • 2 250-mL beakers

  • 1 100-mL graduated cylinder

  • Thermometer

  • Styrofoam Insulated Cup

  • Hot plate, as needed

  • Ice, as needed

PROCEDURE:

  1. For Sample A:

  2. Measure 100 mL of cold tap water in the graduated cylinder.  

  3. Record your volume in Data Table 1. 

  4. Convert your volume into mass.  The density of water is 1.00 g/mL.

  5. Pour the water into a 250-mL beaker. Record the temperature in Data Table 1.

  6. For Sample B: 

  7. Measure 100 mL of hot tap water in the graduated cylinder.  

  8. Record your volume in Data Table 1.

  9. Convert your volume into mass. The density of water is 1.00 g/mL.

  10. Pour into the second 250-mL beaker.  Record the temperature in Data Table 1.

  11. Record your prediction of what the final temperature will be when the two samples are mixed (when they reach thermal equilibrium).

  12. Pour the two beakers of water into the styrofoam insulated cup.

  13. Measure and record the final temperature in Data Table 1.

  14. Repeat three more times using different water volumes and temperatures. 

  15. Clean all materials and pour the water down the sink.

DATA TABLE 1: Mixing Two Water Samples at Different Volumes and Temperatures

----------- Sample a ------------------------ Sample b

volume, mL temp, C Mass, g volume, mL temp, C Mass, g predicted temp measured final temp
Trial 1 100 25 106 100 75 95 48 40
Trial 2 50 26 50 50 59 48.5 42 39
Trial 3 25 25 24 50 50 48.43 37 40
Trial 4 25 25 24 100 72 98 61 59

  1. In each trial, how does the final temperature compare to the temperature of the colder sample?

  2. In each trial, how does the final temperature compare to the temperature of the warmer sample?

  3. In what direction did heat energy flow?

from the hot water to the cold water

  1. Would it be possible to have a final temperature after mixing that is colder than the cold sample or a final temperature that is hotter than the hot sample?

no, due to the property of diffusion

  1. Imagine that you mix 25 g of water at 25 °C with 25 g of water at 65 °C. Predict the final temperature of the sample.  _________  Explain your answer. 

45?

  1. Suppose you mix 75 g of water at 15°C with 25 g of water at 75°C. Predict the final temperature from the choices. 

  2. 30°C B. 45°C C. 60°C

Explain your choice.

The amount of energy needed to raise the temperature of 1 g of a substance 1°C is called its specific heat capacity, c. The specific heat of a substance (also known as specific heat capacity) determines how quickly the temperature of that material will rise or fall when it gains or loses heat energy. 

Think about how the sand feels when you walk barefoot at the beach on a hot day.  Would you expect sand to have a higher or lower specific heat capacity?  That is, does sand hold more heat than water at the same temperature?  Sand has a heat capacity of 0.18 cal/g°C or 0.795 J/g°C.  The heat capacity of sand is much less than water.  This is why sand feels hot to us.  The energy is released from the sand.

The following table shows the specific heat capacity of several substances. 

Substance Specific Heat Capacity (cal/gC) Specific Heat Capacity (J/gC)
Water, H2O (l) 1.00 cal/g°C 4.184 J/g°C
Ocean water 0.94 cal/g°C 3.697 J/g°C
Sand 0.18 cal/g°C 0.795 J/g°C
Air (sea level, dry) 0.24 cal/g°C 1.004 J/g°C
Ice, H2O (s) 0.50 cal/g°C 2.092 J/g°C
Asphalt 0.22 cal/g°C 0.920 J/g°C
Granite 0.17 cal/g°C 0.711 J/g°C
Styrofoam 0.27 cal/g°C 1.130 J/g°C

  1. Which substance listed in the table requires the least amount of energy to raise the temperature of 1 g of the substance at 1°C?

  2. What substance listed in the table requires the most amount of energy to raise the temperature of 1 g of the substance at 1°C?

  3. The intermolecular force in water is hydrogen bonding.  Why would the presence of hydrogen bonding result in liquid water having a high heat capacity? 

The equation Q = mc∆T, where Q is heat in Joules or Calories, m is mass, c is the heat capacity of the substance, and ∆T, is the change in temperature (Tf - Ti), can be used to calculate any variable.

  1. The lowest temperature recorded for Antarctica ice was in 1983 at a temperature of -89.2°C.  Calculate the amount of heat in both calories and Joules to raise the temperature of the ice in Antarctica from -89.2°C to 0°C.  There are 2.4 x 1022 grams of ice in Antarctica.

  2. How much energy in both calories and joules, would it take to heat the ocean by 1°C assuming that the total mass of the ocean is 1.4 x 1024 g?

  3. In Greenland, as ice sheets melt, sand or dirt is left behind.  Based on the specific heat capacity values, which substance, ice or sand would result in more thermal energy in the atmosphere?

Richmann’s law of mixtures describes the final temperature resulting in thermodynamic equilibrium when two bodies with different initial temperatures are brought into contact.

If two bodies with different initial temperatures are brought into contact with each other, the temperatures will become more and more equal. Eventually, thermodynamic equilibrium will be reached. The temperatures have then completely equalized and a common final temperature has been established, which is also referred to as the mixing temperature.

One can observe such an equalization of temperatures, for example, when pouring hot water into a cold glass. While the glass is heated by the hot water, the water cools down on the relatively cold glass. After some time, the different initial temperatures have equalized and the glass has the same temperature as the water inside it. The final temperature lies between these two initial temperatures.

Depending on how much water is poured into the glass, the final temperature is shifted towards higher or lower values. With a larger amount of water, it can be assumed that higher final temperatures result, since more hot water is present and causes greater heating of the glass. The equalization of temperatures can be explained using the particle model. This is discussed in detail in the article Heat and thermodynamic equilibrium.

If the energy flow is equal to q = mc∆T, and when two substances are being mixed the amount of temperature absorbed equals the amount of energy released given the equation +q = -q.  

Remember that ∆T = Tf - Ti.  To derive the equation for Tf we need the initial equation, m1c1(Tf - T1i) = -[m2c2(Tf - T2i)], where the subscript 1 represents the colder substance and the subscript 2 represents the warmer substance.

Simplified,  m1c1Tf  - m1c1T1i  = -m2c2Tf + m2c2T2i

Tf = m1⋅c1⋅T1i+ m2⋅c2⋅T2im1⋅c1+m2⋅c2

17.  Use this derived equation to predict the final temperature of water for the following problems.  Remember the density of liquid water is 1g/mL and the specific heat of water is 4.18 J/goC.

  1. 150.0 mL of 25.0oC water mixed with 75.0 mL of 50.0oC water. (ans. 33.3 oC)

  2. 100.0 mL of 50.0oC water mixed with 75.0 mL of 70.0oC water. (ans. 58.6 oC)

  3. Take one of your measurements from the experiment and calculate the final temperature.  How close was your measurement compared to your calculated final temperature.

  4. One of the biggest glaciers in Iceland is the Vatnajökull glacier, which has a volume of about 3000 km3 and has a density of 917 kg/m3.  When it melts (0.0 0C), the glacial runoff feeds into the Norwegian sea which has a mass of 1.81 x 1022 kg and an average temperature of 7.0 oC.  If all of the glacier melted, what would be the final temperature of the surrounding Norwegian Sea water?  (ans. 6.99 oC)   

Part 2: How is it possible for water below the freezing point of 0oC is in the liquid phase?

In the previous experiment, you observed water below the freezing point in the liquid phase.  How is it possible?  The answer is colligative properties.  Colligative properties are the physical changes that result from adding solute to a solvent in solutions. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles, although do depend on the type of solvent.   The colligative property that you witnessed is called freezing point depression, where the freezing point is lowered due to dissolved solute in solution.  There are other colligative properties, but for our intents and purposes, we will analyze freezing point depression. 

Let's start this discussion the same way we started the others, by defining the normal freezing point. The normal freezing point of a liquid is the temperature at which a liquid becomes a solid at 1 atm. A more specific definition of a freezing point is the temperature at which solid and liquid phases coexist in equilibrium. For water, the freezing point is 0oC.  Let's see if we can figure out why the freezing point is lowered when we add solutes to a solution. We already know that in order to freeze a liquid, we have to lower the temperature. As the temperature lowers, the solution becomes more ordered as it moves toward the solid phase. This is an effect that works against the second law of thermodynamics. In short, entropy (disorder) likes to increase not decrease in the natural scheme of things. So if we have to lower the temperature to a certain point to freeze a pure solvent, when we add a solute we add to the entropy of the system, right? The mixture is more disordered than the pure. This additional amount of entropy must now be overcome to allow the liquid to change phases into a solid (become ordered). This means that the temperature will have to be even lower than before. Thus the addition of any type of solute to a solvent will lower its freezing point.

The equation to figure out how great the change in temperature will be is given by:

ΔT = iKfm

ΔT= change in temperature

i = the van't Hoff factor, which is the number of particles into which the solute dissociates

m = the molality, which is the moles of solute per kilograms (mols/kg) of solvent

Kf = the molal freezing point constant (for water, Kf = -1.86 oC/m)

So the change in temperature is dependent on the number of particles (i) in one mole of solute dissolved in the amount of solvent (m). 

Let’s do some practice…

  1. What is the freezing point of a solution of 15.0 g of NaCl in 250 g of water? The molal freezing point constant, Kf, for water is -1.86 oC kg / mol.

Ans. -3.79oC

  1. What is the freezing point of a solution of 25.0 g of CaCl2 in 250 g of water? The molal freezing point constant, Kf, for water is -1.86 oC kg / mol.

Ans. -5.03oC

  1. What is the freezing point of a solution of 35.0 g of AlCl3 in 250 g of water? The molal freezing point constant, Kf, for water is -1.86 oC kg / mol.

Ans. -7.81oC

  1. What is the freezing point of a solution of 15.0 g of Al2(SO4)3 in 250 g of water? The molal freezing point constant, Kf, for water is 1.86 oC kg / mol.

Ans. -1.63oC

NEXT STEPS:

  1. Reflect on today’s question: Why does the ocean hold heat so well?

  2. Reflect on the unit question, how does today’s activity relate to the unit question? Open up the IMT for this unit, complete all boxes for lesson 4.

  3. Make sure all parts of the L7.4 student sheet are complete & complete the check for understanding on Schoology.**

Last update: June 5, 2023
Created: June 5, 2023