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Lesson plan Grade 12 Science

Turnover and Intake Depth at Calderhurst Reservoir

A 60-minute grade 12 lesson using a calorimeter mixing investigation and energy balance to explain reservoir stratification and choose an intake blend.

Objectives

  1. Students will show experimentally that energy lost by warm water equals energy gained by cold water in an insulated cup.
  2. Students will predict the temperature of a blend from the masses and temperatures of its parts.
  3. Students will explain why a small temperature difference keeps a reservoir from mixing until autumn.

Vocabulary

Key words: specific heat, epilimnion, hypolimnion, thermocline, turnover, multi-level intake

Teacher overview

Calderhurst Reservoir is a fictional water supply whose treatment plant draws through a multi-level intake, meaning it can pull raw water from any of three depths. This lesson starts with a calorimeter investigation, has students confirm that the energy one portion of water loses equals the energy the other gains, then applies the same balance to the whole reservoir at autumn turnover. Two misconceptions are worth watching for. The first is that wind mixes a lake whenever it blows, when in fact the density difference between a warm surface layer and a cold bottom layer resists mixing until the surface cools. The second is that a difference of a few degrees is too small to matter, which the density figures in the handout contradict.

Procedure

Hook (5 min): Ask students why a treatment plant would build three intake ports at different depths rather than one. Collect answers on the board without resolving them. Note that Calderhurst's operators change ports in the same weeks every year and that the lesson will explain what they are responding to.

Investigation setup and prediction (10 min): Groups measure out two portions of water of clearly different masses, one warm and one cold, and record the mass and starting temperature of each. Before mixing, each group writes a predicted final temperature and a one-sentence reason. Require the prediction in writing first, since a group that mixes before predicting has run a demonstration rather than a test.

Conducting the investigation (12 min): Groups pour both portions into a lidded foam cup, stir gently, and record the temperature every 15 seconds until it stops changing. They record the steady value as the measured final temperature. Groups pour the warm water slowly and wipe up spills right away to avoid scalds and slips.

Class data and the energy balance (8 min): Collect predicted and measured final temperatures on the board. Introduce the specific heat of liquid water, 4.18 joules per gram per degree Celsius, and compute for one group's data the energy released by the warm portion and the energy absorbed by the cold portion. Point out that the two figures agree within the accuracy of the thermometers, which is the evidence that energy moved from one portion to the other rather than disappearing.

From the cup to the reservoir (10 min): Project a temperature profile for Calderhurst in August: 24.0 degrees Celsius at 10 feet, 16.0 degrees at 30 feet, and 8.0 degrees at 55 feet. Name the three layers, the epilimnion, the thermocline, and the hypolimnion. Explain that warm water is less dense than cold water, so the surface layer floats and the wind cannot fold it under. Explain that once autumn cools the surface, the density difference disappears and the whole column mixes, which is turnover. Show the density values in the handout table and have students say aloud how small the differences are.

Independent practice (10 min): Students complete the handout. Circulate and watch for two errors: using the temperature change of the wrong portion in the energy calculation, and setting up the blend equation with the fractions reversed.

Closure and exit ticket (5 min): As the formative check, each student answers one question on a slip: in which season would the operators be able to draw water of the same temperature from all three ports, and why? Scan for answers naming autumn, winter, or early spring and citing the loss of the density difference.

Turnover and Intake Depth at Calderhurst Reservoir — Grade 12 Science
Name: ____________________________ Date: ______________

Student Handout

Use these values throughout.

  • Specific heat of liquid water: 4.18 joules per gram per degree Celsius
  • Density of pure water: 0.99730 grams per milliliter at 24.0 degrees Celsius, 0.99895 at 16.0 degrees, and 0.99985 at 8.0 degrees
  • One cubic meter equals 1,000,000 milliliters

Part 1, the insulated cup.

  1. A 150 g portion of water at 62.0 degrees Celsius is mixed with a 350 g portion at 14.0 degrees Celsius in an insulated cup. Predict the final temperature of the mixture, in degrees Celsius.

  2. Using your answer to item 1, calculate the energy released by the 150 g portion as it cools, in joules.

  3. Using your answer to item 1, calculate the energy absorbed by the 350 g portion as it warms, in joules.

Part 2, the Calderhurst intake. The fictional Calderhurst Regional Water Authority draws through ports at three depths. In August the 30-foot port reads 16.0 degrees Celsius and the 55-foot port reads 8.0 degrees Celsius.

  1. The plant wants a blended raw water temperature of 13.0 degrees Celsius using only the 30-foot and 55-foot ports. Find the percent of the blend that must come from the 55-foot port.

  2. The plant treats 6.0 million gallons per day. Using your answer to item 4, find how many million gallons per day come from the 55-foot port.

Part 3, why the layers stay put.

  1. Find the mass, in kilograms, of 1.00 cubic meter of reservoir water at 24.0 degrees Celsius.

  2. Find the mass, in kilograms, of 1.00 cubic meter of reservoir water at 8.0 degrees Celsius.

  3. Find the difference between your answers to items 6 and 7, in kilograms.

Part 4, turnover. Just before turnover, the fictional reservoir holds an upper layer of 5.0 x 10^8 kilograms at 14.0 degrees Celsius and a lower layer of 3.0 x 10^8 kilograms at 8.0 degrees Celsius.

  1. Predict the temperature of the whole reservoir after complete mixing, in degrees Celsius, assuming no energy is exchanged with the air.

  2. Using your answer to item 9, calculate the energy released by the upper layer during turnover, in joules.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 28.4 degrees Celsius.
    The final temperature is the mass-weighted mean: 150 times 62.0 gives 9,300, and 350 times 14.0 gives 4,900. Their sum, 14,200, divided by the total mass of 500 g, is 28.4 degrees Celsius.

  2. Answer: 21,067 joules.
    The warm portion cools from 62.0 to 28.4, a change of 33.6 degrees. Energy equals 150 times 4.18 times 33.6, which is 21,067 joules.

  3. Answer: 21,067 joules.
    The cold portion warms from 14.0 to 28.4, a change of 14.4 degrees. Energy equals 350 times 4.18 times 14.4, which is 21,067 joules, matching item 2 as conservation of energy requires.

  4. Answer: 37.5 percent.
    Let x be the fraction from the 55-foot port. Then 8.0x plus 16.0 times the quantity 1 minus x equals 13.0, so 16.0 minus 8.0x equals 13.0 and x equals 0.375, or 37.5 percent.

  5. Answer: 2.25 million gallons per day.
    0.375 times 6.0 million gallons per day equals 2.25 million gallons per day from the deep port, leaving 3.75 million from the 30-foot port.

  6. Answer: 997.30 kilograms.
    0.99730 grams per milliliter times 1,000,000 milliliters equals 997,300 grams, which is 997.30 kilograms.

  7. Answer: 999.85 kilograms.
    0.99985 grams per milliliter times 1,000,000 milliliters equals 999,850 grams, which is 999.85 kilograms.

  8. Answer: 2.55 kilograms.
    999.85 minus 997.30 equals 2.55 kilograms per cubic meter. That difference, about one quarter of one percent of the mass, is enough to hold the warm layer on top through an entire summer of wind.

  9. Answer: 11.75 degrees Celsius.
    The mass-weighted mean is 5.0 times 14.0 plus 3.0 times 8.0, which is 70 plus 24, or 94, divided by the total of 8.0 mass units, giving 11.75 degrees Celsius.

  10. Answer: 4.70 x 10^12 joules.
    The upper layer is 5.0 x 10^11 grams and cools 2.25 degrees, so the energy is 5.0 x 10^11 times 4.18 times 2.25, or about 4.70 x 10^12 joules. The lower layer absorbs the same amount as it warms 3.75 degrees.

Exit ticket answer: Late autumn or winter, because once the surface cools to the temperature of the deep water the density difference vanishes and wind can mix the whole column, so every port draws water of nearly the same temperature. Also accept early spring, when the surface has warmed to about the deep-water temperature and the column mixes before summer layering sets up.

Differentiation

Support

Give these students the mass-weighted mean written as a filled-in template with labeled blanks for mass and temperature, and provide items 6 and 7 with the multiplication set up so only the unit conversion from grams to kilograms remains. Pair them with a partner who reads the profile values aloud.

Extension

Ask students to determine the surface temperature at which turnover would begin if the deep layer held steady at 8.0 degrees Celsius, then explain why water reaching 4 degrees Celsius complicates the simple rule that colder water always sinks.

ELL supports

Pre-teach epilimnion, hypolimnion, thermocline, and turnover with a labeled depth profile showing the three layers and the intake ports drawn at their depths. Provide the handout with the specific heat and density values repeated beside each item that needs them, and allow students to state the exit ticket answer orally.

Extensions

Have students obtain a published depth-temperature profile for a real reservoir in their region across several months and mark the week turnover occurred. Ask them to explain what an operator would watch for in the raw water during that week, given that the bottom layer has spent the summer isolated from the air.