HS-PS3-4: Investigating Thermal Energy Transfer and the Second Law
In plain English: HS-PS3-4 is the NGSS high school performance expectation that asks students to plan and conduct an investigation showing that thermal energy flows from the hotter to the cooler part of a closed system until the temperatures even out. Students mix water or add warm metal to water, track the temperatures and compare the energy lost with the energy gained. It is usually taught in Physics, Chemistry or Physical Science.
Plan and conduct an investigation to provide evidence that the transfer of thermal energy when two components of different temperature are combined within a closed system results in a more uniform energy distribution among the components in the system (second law of thermodynamics).
Clarification Statement: Emphasis is on analyzing data from student investigations and using mathematical thinking to describe the energy changes both quantitatively and conceptually. Examples of investigations could include mixing liquids at different initial temperatures or adding objects at different temperatures to water.
Assessment Boundary: Assessment is limited to investigations based on materials and tools provided to students
Next Generation Science Standards · Disciplinary Core Idea: Energy · Official standard
Students plan and conduct investigations in which two parts of a nearly closed system start at different temperatures: hot and cold water mixed in a foam cup, hot water in a can sitting in cold water, and warm metal dropped into water. They measure how the temperatures change, compute the thermal energy each part lost or gained with Q = mcΔT, and use the data as evidence that energy always flows from the hotter part to the cooler part until the energy is spread more evenly. This one-way spreading is what the standard calls the second law of thermodynamics.
The NGSS Clarification Statement says: "Emphasis is on analyzing data from student investigations and using mathematical thinking to describe the energy changes both quantitatively and conceptually. Examples of investigations could include mixing liquids at different initial temperatures or adding objects at different temperatures to water." The Assessment Boundary says: "Assessment is limited to investigations based on materials and tools provided to students". Every investigation on this page uses only the classroom kit listed in Materials. All data sets are invented for teaching. The page uses specific heats of 4186 J/(kg·°C) for water, 900 for aluminum, 450 for steel and 385 for copper.
Learning Objectives
By the end of this lesson, students will be able to:
Plan an investigation of two components at different temperatures in a nearly closed system, naming the variables, controls, measurements and safety steps
Conduct the investigation with the provided tools and record temperature data for both components
Use Q = mcΔT to compare the thermal energy lost by the hotter component with the energy gained by the cooler one, and explain any gap
Predict the final temperature of a mixture and test the prediction with data
Use temperature-time data as evidence that thermal energy spreads from hot to cold until the distribution is more uniform (second law of thermodynamics)
Prior Knowledge Required
Students should already be comfortable with:
Temperature as a measure of the average kinetic energy of particles MS-PS1-4
Energy transfer depends on the type of matter, the mass and the temperature change MS-PS3-4
Reading a thermometer and a balance, and drawing a line graph from a data table
Solving a linear equation for one unknown HSA.REI.B.3
Show a mug of hot cocoa and a small jug of cold milk. Ask students to predict before any numbers appear:
Warm-Up Prompt
"You pour cold milk into hot cocoa and stir. Sketch the temperature of the cocoa and of the milk against time on one graph. Will the mixture ever separate back into a hot part and a cold part on its own? What would you measure to find out?"
Collect the sketches. Many students draw both lines meeting at a middle temperature; ask whether the middle is always halfway. Record their measurement ideas (masses, starting temperatures, final temperature, time) on the board.
Direct Instruction15-20 minutes
Closed systems and the direction of energy flow. In this standard, a closed system means one in which no energy enters or leaves (physicists call this an isolated system; in physics and chemistry, "closed" often means only that no matter enters or leaves). Nested foam cups with a lid come close. When two components at different temperatures are in contact, thermal energy flows from the hotter to the cooler one, never the other way on its own, until they reach the same temperature. The energy a component gains or loses is Q = mcΔT. In a closed system the energy lost by the hot part equals the energy gained by the cold part, so the total stays the same; what changes is how evenly it is spread. Show the can-in-a-cup setup in Diagram 2.
Equal masses of water
0.150 kg of water at 60 °C is mixed with 0.150 kg of water at 20 °C in a foam cup with a lid.
Result: Equal masses of the same substance meet halfway: final temperature 40 °C. Energy lost by the hot water = 0.150 × 4186 × 20 = 12,558 J, equal to the energy gained by the cold water.
Unequal masses of water
0.100 kg of water at 58 °C is mixed with 0.300 kg of water at 10 °C.
Result: Final temperature = (0.100 × 58 + 0.300 × 10) ÷ 0.400 = 22 °C, closer to the larger mass. Hot water loses 0.100 × 4186 × 36 = 15,069.6 J; cold water gains 0.300 × 4186 × 12 = 15,069.6 J.
Warm metal added to water
A 0.200 kg aluminum block taken from an 80 °C water bath is placed in 0.250 kg of water at 20 °C.
Result: Energy lost by the block = energy gained by the water: 0.200 × 900 × (80 - T) = 0.250 × 4186 × (T - 20), so T ≈ 28.8 °C. Water's large specific heat keeps the final temperature near the water's start.
Realistic lab data (invented) with a small gap
0.200 kg of water at 55.0 °C is mixed with 0.200 kg at 15.0 °C. The prediction is 35.0 °C, but the thermometer reads 34.2 °C.
Result: Hot lost 0.200 × 4186 × 20.8 = 17,413.8 J; cold gained 0.200 × 4186 × 19.2 = 16,074.2 J. The missing 1,339.5 J (about 7.7%) went into the cup and the air: the system was nearly, not perfectly, closed.
After each example, ask: "Which way did the energy flow, how do you know, and does the total add up?" Stress that temperature and thermal energy are different: the 0.300 kg of cold water in the second example gains as much energy as the hot water loses but changes temperature by only a third as much.
Guided Practice15 minutes
Pairs analyze invented data from the can-in-a-cup setup in Diagram 2: 0.12 kg of hot water in a thin aluminum can sits in 0.36 kg of cold water inside nested foam cups with a lid. Two probes record both temperatures. Then compare with Diagram 1.
Guided practice data: can-in-a-cup (invented)
t (min)
0
1
2
3
4
6
8
10
Hot water (°C)
60.0
45.8
37.2
32.0
28.9
25.8
24.7
24.2
Cold water (°C)
12.0
16.7
19.6
21.3
22.4
23.4
23.8
23.9
Compute the temperature difference at each time. It starts at 48.0 °C. How does it change, and does it ever change sign?
At t = 4 min, compute the energy lost by the hot water, 0.12 × 4186 × (60.0 - 28.9), and the energy gained by the cold water, 0.36 × 4186 × (22.4 - 12.0). Are they close?
Predict the final common temperature from the masses and starting temperatures, and check it against the last readings.
Write one sentence that uses the data as evidence that the energy distribution became more uniform.
Independent Practice10-15 minutes
Each student writes a one-page plan for a metal-in-water investigation using only the kit: the question, the variables (changed, measured, kept the same), the steps, a data table, the number of trials and the safety steps. The teacher heats the metal samples in a water bath; students move them with tongs.
Students then analyze one invented trial: a 0.150 kg copper sample at 75.0 °C is placed in 0.200 kg of water at 18.0 °C, and the final temperature is 21.6 °C. They find the energy lost by the copper (about 3,084 J), the energy gained by the water (about 3,014 J), the predicted final temperature (about 21.7 °C) and explain what the small gap means.
Closure5 minutes
Exit ticket: "You mix 0.050 kg of water at 60 °C with 0.150 kg of water at 16 °C. Predict the final temperature and explain which way the energy flows and why the mixture will not separate again." (27 °C.) Return to the warm-up sketches and ask students to mark the predicted meeting temperature on each.
Differentiation Strategies
For Struggling Students
Give a data table with columns for m, c, starting temperature, final temperature, ΔT and Q, so students compute one column at a time
Start with equal masses of water before unequal masses and metals, and use whole-number temperatures
Provide a claim-evidence-reasoning frame: "The hot part lost ___ J and the cold part gained ___ J. The temperatures moved toward ___. This shows ___."
For Advanced Students
Ask students to use the energy balance to find the specific heat of an unknown metal sample from the kit and compare it with a table of values
Have students estimate how much energy the cup absorbs in a trial and design a way to measure it
Ask students to explain, using the guided-practice data, why the temperature difference shrinks quickly at first and slowly later
Assessment Guidance
What to Look For
Check that plans name one changed variable, the measured temperatures and the controls (masses, starting temperatures, cup, lid, stirring). Look for data tables that record both components, not only the final temperature. Check that students compute energy with mass and specific heat, not with temperature alone, and that they explain gaps between energy lost and gained by leaks to the cup and air rather than by energy disappearing. Listen for students who say "cold flows into the warm object".
02
Classroom Activities
3 Activities
1
Mixing Water Lab: Predict, Mix, Measure
30 minGroups of 3
Groups plan and conduct three mixtures of hot and cold water in nested foam cups with a lid. For each mixture they predict the final temperature from the masses and starting temperatures, mix, stir, and record the final reading.
Procedure
Plan first: write the question, the variable you will change (the mass of hot or of cold water), what you keep the same and your data table. Get the plan checked before you start.
Measure each mass of water on the balance in a cup. Record each starting temperature just before mixing.
Pour the hot water into the cold, put on the lid, stir gently and record the highest steady reading.
Safety: wear goggles, keep hot water at or below 60 °C, carry cups with two hands, and wipe up spills at once.
Sample Data (invented)
Trial 1: 0.100 kg at 50.0 °C with 0.100 kg at 20.0 °C, predicted 35.0 °C, measured 34.4 °C. Trial 2: 0.100 kg at 50.0 °C with 0.200 kg at 20.0 °C, predicted 30.0 °C, measured 29.6 °C. Trial 3: 0.200 kg at 50.0 °C with 0.100 kg at 20.0 °C, predicted 40.0 °C, measured 39.1 °C.
Discussion Questions
In all three trials the measured temperature is a little below the prediction. What does that tell you about where some energy went?
The gap is largest in Trial 3, which has the most hot water. Why might more hot water mean a larger gap?
How do the data show that the final temperature lies closer to the starting temperature of the larger mass?
Modification for Distance Learning
Students run the investigation at home with two mugs, a kitchen scale and a kitchen thermometer (hot tap water only), or analyze the sample data above, and post their prediction table to the class board.
2
Can-in-a-Cup: Watching Temperatures Converge
25 minPairs
Pairs set up the apparatus in Diagram 2 so that the hot and cold water exchange energy through the can wall without mixing. They record both temperatures every minute for 10 minutes, graph them, and compare their graph with Diagram 1.
Procedure
Measure the masses of the hot water (in the can) and the cold water (in the cup) on the balance. Choose a mass ratio different from the guided-practice data.
Place the can in the cup, fit the lid and probes, and start the timer. Stir the hot water in the can gently every 30 s.
Record both temperatures every minute. After 10 minutes, compute the energy lost by the hot water and gained by the cold water.
Safety: goggles, hot water at or below 60 °C, and keep probe cables away from the table edge.
Discussion Questions
Did the temperature difference ever grow during your run? What would it mean if it did?
Compare your energy lost and energy gained. What share of the energy lost is missing, and where could it be?
Leave a control cup of room-temperature water next to your setup. Does part of it ever become hotter on its own? What does that show about the direction of energy flow?
Challenge Variation
Pairs predict, then test, how the graph changes if the aluminum can is replaced by a thick plastic cup of the same size. They explain the prediction in terms of how fast energy passes through the wall and whether the final temperature should change.
3
Plan and Data Critique Stations
20 minGroups of 3-4
Four stations each show a student investigation plan or result, all invented. Groups decide whether it gives good evidence for the second law, name any flaw and write a fix. Groups rotate every 5 minutes.
Station Cards
Station 1: "Mix 0.100 kg of 50 °C water with 0.100 kg of 20 °C water in an open glass beaker and read the thermometer once, 10 minutes later."
Station 2: "Trial 1: 0.100 kg of hot water at 40 °C. Trial 2: 0.200 kg of hot water at 55 °C. Compare the final temperatures to see how the mass of hot water matters."
Station 3: "We mixed equal masses of water at 30 °C and 50 °C and measured 52 °C, so mixing adds energy."
Station 4: "Two foam cups of water at 25.0 °C sat side by side with their walls touching for 20 minutes. Both still read 25.0 °C."
Discussion Questions
At Station 1, name two ways energy leaves the system and one change to the plan that reduces each.
At Station 2, which two variables change between trials, and how would you fix the plan?
At Station 3, why is 52 °C impossible for this mixture, what should the reading have been, and what error could explain the result?
Station 4 shows no change at all. How is that also evidence about the direction of thermal energy flow?
Modification for Distance Learning
Post the four cards as slides. Groups meet in breakout rooms and submit one critique per station on a shared document.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Two Temperatures Converging in a Closed System
The guided-practice data drawn to scale. The hot water (red) cools and the cold water (blue) warms until both approach 24 °C, the temperature predicted by the energy balance. The gap between the lines only shrinks: thermal energy flows one way, from hot to cold, and ends up spread more evenly.
Diagram 2: The Can-in-a-Cup Setup
A schematic, not to scale. Hot water in a thin aluminum can sits in cold water inside two nested foam cups with a lid. Thermal energy passes through the can wall, the foam and lid keep most of it inside the system, and two probes record both temperatures so students see the whole transfer, not only the end.
04
Homework Assignment
~30 min
HS-PS3-4 Homework: Planning and Analyzing Thermal Energy Investigations
Directions: All data sets are invented for practice, and every investigation uses only the classroom kit (foam cups and lids, thin aluminum cans, probes reading to 0.1 °C, a balance, hot water at most 60 °C, and metal samples heated by the teacher). Use specific heats of 4186 J/(kg·°C) for water and 450 J/(kg·°C) for steel. Show each calculation with units, and end each problem with a sentence that names your evidence.
Part 1: Predicting and Checking Final Temperatures (Problems 1-2)
You mix 0.300 kg of water at 58 °C with 0.100 kg of water at 14 °C in a closed foam cup. (a) Predict the final temperature. (b) Compute the energy lost by the hot water and the energy gained by the cold water. (c) Explain why the final temperature is not halfway between 58 °C and 14 °C.
A 0.180 kg steel sample from an 80 °C water bath is placed in 0.200 kg of water at 20 °C. (a) Predict the final temperature. (b) The measured final temperature is 25.0 °C. Using this reading, compute the energy lost by the steel and gained by the water. (c) What percent of the energy lost by the steel did the water gain, and where did the rest go?
Part 2: Planning an Investigation (Problems 3-4)
Plan an investigation, using only the kit, to test this question: "Does the final temperature of a water mixture depend on the ratio of hot water mass to cold water mass?" Name the variable you change, the variable you measure, three variables you keep the same, the data table, the number of trials and two safety steps.
A student plans to drop 0.050 kg of water at 30 °C into 0.450 kg of water at 20 °C and read the result with a thermometer marked in whole degrees. (a) Predict the final temperature. (b) Explain why this plan gives weak evidence. (c) Rewrite the plan with the kit so the temperature changes are easy to measure.
Part 3: Analyzing Data and Arguing from Evidence (Problems 5-6)
In a can-in-a-cup run, 0.080 kg of hot water sits in 0.240 kg of cold water. The readings at t = 0, 2, 4, 6 and 8 min are: hot 58.0, 43.4, 35.9, 32.1, 30.1 °C; cold 18.0, 22.9, 25.4, 26.6, 27.3 °C. (a) Compute the temperature difference at each time. (b) At t = 4 min, compute the energy lost by the hot water and gained by the cold water. (c) Predict the common final temperature. (d) Explain how the data support the second law of thermodynamics.
A student says: "When 0.200 kg of water at 40 °C is mixed with 0.200 kg at 20 °C, the energy of the system goes down, because the final temperature is lower than 40 °C." Use calculations of the energy lost and gained to write a claim, evidence and reasoning response. Say what does change during the mixing.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Energy Calculations
Q = mcΔT used correctly with units for both components, and energy lost compared with energy gained
Correct method with one calculation or unit error
Investigation Plan
One changed variable, both temperatures measured, clear controls, repeated trials and safety steps, all with kit materials
Plan works but misses controls, repeats or safety
Evidence and Reasoning
Claim cites specific data (a shrinking temperature difference, matching energies) and links them to one-way energy flow
Claim stated but the evidence is general or the link to the second law is missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All data sets are invented, every system is a foam cup with a lid unless stated otherwise, and the specific heat of water is 4186 J/(kg·°C). Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A warm steel block is placed in cool water in an insulated, lidded cup. What happens?
Answer: B
Thermal energy moves from the hotter object to the cooler one until their temperatures are equal. Choice A describes "cold" as a thing that flows, a common misconception; only energy is transferred. Choice C ignores the temperature difference that drives the transfer. Choice D describes energy leaving the system, which the insulated cup largely prevents.
Question 2 of 20 · Multiple Choice
0.200 kg of water at 60 °C is mixed with 0.200 kg of water at 30 °C. What is the final temperature?
Answer: A
Equal masses of the same substance meet halfway: (60 + 30) ÷ 2 = 45 °C. Choice B adds the temperatures. Choice C assumes the hot water cools all the way to the cold water's start, and choice D takes half of the temperature difference instead of the midpoint.
Question 3 of 20 · Multiple Choice
0.100 kg of water at 60 °C is mixed with 0.300 kg of water at 24 °C. What is the final temperature?
Answer: C
Energy lost equals energy gained: 0.100(60 - T) = 0.300(T - 24), so T = (6 + 7.2) ÷ 0.400 = 33 °C, closer to the larger mass. Choice A is the simple average, which ignores the masses. Choice B swaps the masses. Choice D adds the temperatures.
Question 4 of 20 · Multiple Choice
Hot water in a thin can sits in cold water, and both temperatures are recorded for 10 minutes. Which description of the graph is supported by the second law?
Answer: D
Energy flows from hot to cold until the temperatures are equal, so the lines approach each other and level off together. Choice A ignores the energy the cold water gains. Choice C would need energy to flow from cold to hot on its own, which the second law rules out.
Question 5 of 20 · Multiple Choice
0.350 kg of water in a foam cup cools by 12 °C. How much thermal energy did it lose?
Answer: A
Q = mcΔT = 0.350 × 4186 × 12 = 17,581.2 J. Choice B multiplies only the mass and the temperature change. Choice C leaves out the temperature change, and choice D leaves out the mass.
Question 6 of 20 · Multiple Choice
Which observation is the best evidence that the energy in a two-part system has become more evenly distributed?
Answer: B
Equal, steady temperatures in both parts, after a large starting difference, show that the transfer has run until the energy is spread evenly. Choice A shows transfer has started but not where it ends. Choices C and D show energy leaking to the surroundings, which is a flaw, not the evidence the standard asks for.
Question 7 of 20 · Multiple Choice
You want to test how the mass of hot water affects the final temperature of a mixture. Which variable must you keep the same in every trial?
Answer: C
A fair test changes only the hot water mass, so both starting temperatures (and the cold water mass) must stay the same. Choice A is the variable being changed, and choice B is the variable being measured.
Question 8 of 20 · Multiple Choice
Why do investigations for HS-PS3-4 use two nested foam cups with a lid?
Answer: D
Foam and a lid slow the transfer of energy to the air and the table, so the energy lost by the hot part nearly equals the energy gained by the cold part. Choices A and B describe effects the cups do not have, and choice C is wrong because mixing is the point of the investigation.
Question 9 of 20 · Multiple Choice
In one trial, the hot water lost 9,400 J and the cold water gained 8,700 J. Which explanation is best?
Answer: B
The 700 J difference is energy that left the two waters for the surroundings, since no system in a classroom is perfectly closed. Choice A contradicts conservation of energy. Choice D reverses the direction of the transfer, which would make the cold water lose energy, not gain it.
Question 10 of 20 · Multiple Choice
A 0.100 kg copper sample (c = 385 J/(kg·°C)) at 80 °C is placed in 0.100 kg of water at 20 °C. What is the final temperature, to the nearest degree?
Answer: A
0.100 × 385 × (80 - T) = 0.100 × 4186 × (T - 20) gives T = 11,452 ÷ 457.1 ≈ 25 °C. Water's much larger specific heat keeps the final temperature near 20 °C. Choice B is the simple average, which ignores the specific heats. Choice C swaps the specific heats. Choice D assumes the water does not warm at all.
Question 11 of 20 · Multiple Choice
Two metal blocks, both at 40 °C, touch inside an insulated box. What happens next?
Answer: C
Net thermal energy transfer needs a temperature difference, and the blocks have none. Choices A and B confuse the total thermal energy (larger for the larger block) with temperature, which decides the direction of flow. Choice D would need energy to leave the insulated box.
Question 12 of 20 · Multiple Choice
A student says: "When hot and cold water mix, energy disappears, because the hot water ends up cooler." What is the best response?
Answer: D
In a closed system the energy lost by the hot water is gained by the cold water, so the total is unchanged. What changes is the distribution. Choice B confuses energy leaving the system with energy disappearing, and choice C gets the direction wrong: the cold water gains energy.
Question 13 of 20 · Multiple Choice
Which result, in an insulated box, would contradict the second law of thermodynamics?
Answer: B
Energy flowing on its own from the cooler block to the hotter block makes the distribution less uniform, which the second law rules out. Choices A and C both show the temperatures moving toward each other; the final value depends on the masses and materials. Choice D shows no transfer without a temperature difference.
Question 14 of 20 · Multiple Choice
Why should students stir a mixture gently before reading its final temperature?
Answer: C
Without stirring, warm and cool layers can sit in the cup, and one probe may read a local value. Choice A overstates a tiny effect: gentle stirring adds a negligible amount of energy. Choice B describes energy leaving the system, which is not the purpose.
Question 15 of 20 · Short Answer
You mix 0.120 kg of water at 57 °C with 0.280 kg of water at 15 °C in a closed foam cup. Predict the final temperature and show that the energy lost equals the energy gained.
T = (0.120 × 57 + 0.280 × 15) ÷ 0.400 = (6.84 + 4.2) ÷ 0.400 = 27.6 °C. Lost: 0.120 × 4186 × 29.4 = 14,768.2 J. Gained: 0.280 × 4186 × 12.6 = 14,768.2 J. They are equal, as they must be in a closed system.
Question 16 of 20 · Short Answer
Plan an investigation with the kit to show that adding a 0.100 kg aluminum sample at 60 °C to cool water gives a more uniform energy distribution. Name what you measure, what you keep the same and one safety step.
Sample plan: measure the mass of the water (for example 0.200 kg) and the starting temperatures of both the sample and the water, then the water temperature every 30 s after the sample goes in, until it is steady. Keep the cup, lid, water mass, starting temperatures and stirring the same across repeated trials. Evidence: the sample's and the water's temperatures end equal, and the energy the water gains is close to 0.100 × 900 × (60 - Tfinal). Safety: the teacher heats the sample; students move it with tongs and wear goggles.
Question 17 of 20 · Short Answer
0.150 kg of water at 58.0 °C is mixed with 0.150 kg at 18.0 °C. The prediction is 38.0 °C, but the reading is 37.4 °C. Compute the energy lost and gained, the gap as a percent of the energy lost, and explain the gap.
Lost: 0.150 × 4186 × 20.6 = 12,934.7 J. Gained: 0.150 × 4186 × 19.4 = 12,181.3 J. Gap: 753.5 J, about 5.8% of the energy lost. The gap is energy that went into the cup, the lid and the air: the system is only nearly closed. It does not mean energy was destroyed.
Question 18 of 20 · Short Answer
In a can-in-a-cup run, the temperature difference between the hot and cold water is 40, 20, 10 and 5 °C at t = 0, 2, 4 and 6 min. Predict the difference at 8 min and explain what the pattern shows.
The difference halves every 2 min, so at 8 min it is about 2.5 °C. The difference shrinks quickly while it is large and more slowly as it gets small, but it never grows or changes sign: energy keeps flowing one way, from hot to cold, until the distribution is uniform.
Question 19 of 20 · Short Answer
A mixture of hot and cold water sits in an open glass beaker for an hour and ends at room temperature, 22 °C. Is this a closed system? Explain how the result still fits the second law.
It is not closed: the glass and the open top let energy pass to the room. The result still fits the second law, because energy kept flowing from the warmer water to the cooler room until the water and the room were at the same temperature. The energy spread over a much larger system, the room, which is an even more uniform distribution.
Question 20 of 20 · Short Answer
A student mixes 0.300 kg of water at 50 °C with 0.100 kg at 14 °C and predicts 32 °C. Explain the error and give the correct prediction.
The student took the simple average, which only works for equal masses. With three times as much hot water, the final temperature is closer to 50 °C: T = (0.300 × 50 + 0.100 × 14) ÷ 0.400 = 41 °C. Check: the hot water loses 0.300 × 4186 × 9 J and the cold water gains 0.100 × 4186 × 27 J, which are equal.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-PS3-4 mean?
HS-PS3-4 asks students to plan and conduct an investigation showing that when two parts of a closed system start at different temperatures, thermal energy flows from hot to cold until the energy is spread more evenly. That one-way spreading is the second law of thermodynamics. The code means High School, Physical Science, core idea 3 (Energy), performance expectation 4.
Is HS-PS3-4 taught in physics or chemistry?
It is usually taught in Physics or Physical Science, and it fits well in Chemistry units on thermochemistry and calorimetry. The math is Q = mcΔT and solving a linear equation for the final temperature, so it suits students in or after Algebra I.
What is the assessment boundary for HS-PS3-4?
Assessment is limited to investigations based on materials and tools provided to students. So the investigations use a classroom kit (foam cups, thermometers or probes, a balance, water and metal samples) rather than equipment students must find or build themselves. Formal entropy calculations are beyond this standard.
What investigations work for HS-PS3-4?
The Clarification Statement names mixing liquids at different starting temperatures and adding objects at different temperatures to water. A can of hot water inside a cup of cold water is a useful variation because students can track both temperatures as they converge. All three need only foam cups, a balance and a thermometer.
How do you predict the final temperature when mixing water?
Set the energy lost by the hot water equal to the energy gained by the cold water and solve for the final temperature. For water with water, the specific heat cancels, so the final temperature is the mass-weighted average. For example, 0.20 kg at 45 °C and 0.60 kg at 25 °C give (0.20 × 45 + 0.60 × 25) ÷ 0.80 = 30 °C.
What does "more uniform energy distribution" mean in HS-PS3-4?
It means the thermal energy ends up spread evenly instead of concentrated in the hot part. At the start, the particles of the hot component have a higher average kinetic energy than those of the cold one. After the transfer, the temperatures are equal, so the average kinetic energy of the particles is the same in both parts.
How is this investigation about the second law of thermodynamics?
The second law says that on its own, thermal energy flows from hotter to cooler objects and not the reverse, so energy spreads out over time. The investigation shows exactly that: the temperature difference only shrinks, and a mixture never separates back into hot and cold parts. The idea of entropy formalizes this, which is beyond this standard.
Why don't the energy lost and the energy gained match in our lab?
A foam cup is nearly, not perfectly, closed. Some energy goes into the cup, the lid, the thermometer and the air, so the cold water usually gains a little less than the hot water loses. A gap of a few percent up to about 10% is normal for classroom equipment. Ask students which way each source of error pushes the result.
Is heat the same as temperature?
No. Temperature measures the average kinetic energy of the particles, while thermal energy transferred (heat) depends on the mass and the material as well. A bathtub of warm water holds far more thermal energy than a cup of hotter water, but the cup has the higher temperature, and the temperature decides which way the energy flows.
How do you keep a thermal energy lab safe?
Use hot water at or below 60 °C for student handling, let the teacher heat metal samples in a water bath, and have students move hot samples with tongs. Students wear goggles, carry cups with two hands and use foam cups that do not get hot on the outside. No open flames are needed for these investigations.
07
Related Standards
5 standards
These standards connect to HS-PS3-4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-PS3-4Prerequisite
Plan an investigation of how energy transfer relates to type of matter, mass and temperature change
NGSS is a registered trademark of WestEd. Neither WestEd nor the lead states and partners that developed the Next Generation Science Standards were involved in the production of this product, and do not endorse it.
