D. Energy and Society · Thermal and Nuclear EnergyThermal Equilibrium, Mixtures and Measuring Specific Heat Capacity
D2.9 D2.10 D2.1 A1.5 A1.8 A1.13About 75 minutes
Learning goals- I can apply conservation of energy to mixtures: heat lost by the warmer substance equals heat gained by the cooler one.
- I can calculate the final temperature when two samples of water are mixed.
- I can determine the specific heat capacity of a substance by the method of mixtures.
- I can calculate percent error and identify sources of error in a calorimetry investigation.
Words to know
calorimeter
An insulated container used to measure heat transfer, such as two nested foam cups with a lid and a thermometer.
calorimetry
Measuring heat transfer by measuring temperature changes in an insulated system.
method of mixtures
Finding an unknown quantity (such as a specific heat capacity) by mixing a hot and a cold substance and using heat lost = heat gained.
isolated system
A system that exchanges no energy with its surroundings. A good calorimeter comes close to this.
percent error
|experimental value − accepted value| ÷ accepted value × 100%.
The perfect bath
You run a bath that is too hot, so you add cold water. Without a thermometer you test it with your hand and keep adjusting. Physics can do better: if you know the masses and the starting temperatures, you can predict the final temperature before the water even mixes.
The idea is simple and powerful. When a hot thing and a cold thing are mixed in an insulated container, energy is conserved. Every joule the hot water loses is gained by the cold water. In this lesson you will use that idea to predict final temperatures and to measure the specific heat capacity of a substance, the way physics students do in the lab.
Heat lost = heat gained
When a hot sample and a cold sample are placed in contact in an isolated system (no energy enters or leaves), heat flows from hot to cold until they reach the same final temperature T_f. Conservation of energy says:
heat lost by the hot sample = heat gained by the cold sample
With the sign convention from the last lesson (lost heat is negative), the same idea is written:
Q_hot + Q_cold = 0, that is, m₁c₁(T_f − T₁) + m₂c₂(T_f − T₂) = 0
Both forms give the same answer. The second form is handy because the signs take care of themselves: you just solve for the one unknown.
A real container is never perfectly isolated. In the lab we use a calorimeter: two nested foam cups with a lid. Foam traps air, so it is a poor conductor; the lid cuts convection and evaporation. Then we can assume that no heat is lost, and estimate how much error that assumption causes.
Worked exampleExample 1: Mixing two samples of water
0.200 kg of water at 80 °C is mixed with 0.300 kg of water at 20 °C in an insulated container. Find the final temperature.
- Both samples are water, so c is the same for both. Write Q_hot + Q_cold = 0.
- (0.200)(4180)(T_f − 80) + (0.300)(4180)(T_f − 20) = 0
- Divide every term by 4180: 0.200(T_f − 80) + 0.300(T_f − 20) = 0
- Expand: 0.200T_f − 16 + 0.300T_f − 6 = 0, so 0.500T_f = 22.
- T_f = 22 ÷ 0.500 = 44 °C.
- Check: hot water loses (0.200)(4180)(36) = 30 096 J; cold water gains (0.300)(4180)(24) = 30 096 J. Equal, as they should be.
The final temperature is 44 °C. It is closer to 20 °C because there was more cold water.
In Example 1 the hot water drops 36 °C and the cold water rises 24 °C. The products mass × ΔT are equal (0.200 × 36 = 0.300 × 24 = 7.2), because both samples are water.Shortcut for water + water
When two samples of the same substance are mixed, c cancels, and the final temperature is the mass-weighted average: T_f = (m₁T₁ + m₂T₂) ÷ (m₁ + m₂). This shortcut does not work for two different substances.
Measuring specific heat capacity: the method of mixtures
Here is how a physics class determines the specific heat capacity of a metal (expectation D2.9).
- Measure the mass of the metal sample and the mass of cold water in the calorimeter.
- Heat the metal in a beaker of boiling water for several minutes so it reaches about 100 °C. Measure that water's temperature; it is the metal's starting temperature.
- Measure the starting temperature of the cold water in the calorimeter.
- Quickly transfer the metal into the calorimeter, put on the lid, stir gently, and record the highest temperature reached. That is T_f.
- Use heat lost by metal = heat gained by water and solve for c(metal).
Safety in the lab: boiling water and hot metal cause burns. Use tongs, wear safety glasses, keep cords away from water, and let the teacher handle the hot plate if required.
Sources of error: heat lost to the air and to the cup while transferring the metal; hot water clinging to the metal; heat absorbed by the thermometer and cup; reading the thermometer before it reaches its maximum; the metal not fully reaching 100 °C.
Worked exampleExample 2: Identify the metal
A 0.150 kg metal cylinder is heated to 100.0 °C and placed in a calorimeter holding 0.200 kg of water at 20.0 °C. The final temperature is 25.0 °C. (a) Find the specific heat capacity of the metal. (b) The metal is believed to be copper (accepted c = 390 J/(kg·°C)). Find the percent error.
- Heat gained by water: Q = (0.200)(4180)(25.0 − 20.0) = 4180 J.
- Heat lost by metal: (0.150)c(100.0 − 25.0) = 11.25c (in joules).
- Heat lost = heat gained: 11.25c = 4180, so c = 4180 ÷ 11.25 = 372 J/(kg·°C) (to 3 significant digits).
- (b) Percent error = |371.6 − 390| ÷ 390 × 100% = 18.4 ÷ 390 × 100% ≈ 4.7% (using the unrounded value of c).
- The experimental value is low. That makes sense: some heat escaped to the air and cup during the transfer, so the water gained less heat than the metal really lost, which makes the calculated c too small.
c ≈ 3.72 × 10² J/(kg·°C), a 4.7% error from copper's accepted value.
Worked exampleExample 3: Two different substances
A 0.50 kg aluminium block at 90 °C is dropped into 1.0 kg of water at 15 °C. Find the final temperature. (c(Al) = 900, c(water) = 4180 J/(kg·°C))
- Q_Al + Q_water = 0: (0.50)(900)(T_f − 90) + (1.0)(4180)(T_f − 15) = 0
- 450T_f − 40 500 + 4180T_f − 62 700 = 0
- 4630T_f = 103 200
- T_f = 103 200 ÷ 4630 ≈ 22 °C.
- Reasonable? Yes: water has more mass and a much higher c, so the final temperature should be much closer to the water's starting temperature.
Just for funWhy did the hot water and the cold water finally stop arguing?
They reached an agreement at the same temperature. (Physicists call it thermal equilibrium; families call it bath time.)
Hands-on activityKitchen calorimetry
You need: a kitchen scale, a kitchen thermometer, two insulated mugs or foam cups with a lid or plate on top, hot and cold tap water, and about 300 g of stainless steel cutlery (forks and spoons). Grown-up note: hot tap water can scald; a grown-up should handle it.
Part A: predict and test a mixture.
- Measure 150 g of cold tap water into mug 1 and 100 g of hot tap water into mug 2. Measure both temperatures.
- Predict T_f with T_f = (m₁T₁ + m₂T₂) ÷ (m₁ + m₂).
- Pour mug 2 into mug 1, cover, stir and record the steady temperature. Calculate the percent difference from your prediction.
Part B: estimate the specific heat capacity of steel.
- Weigh the cutlery. Stand it in a jug of hot tap water for 5 minutes. Measure the hot water's temperature (this is the cutlery's starting temperature).
- Put 200 g of cold water in a clean insulated mug and measure its temperature.
- Quickly move the cutlery (shake off drops) into the cold water, cover, stir and record the highest temperature.
- Use heat lost by steel = heat gained by water to calculate c(steel). Accepted values for stainless steel are roughly 500 J/(kg·°C). Find your percent error.
What to look for: in Part A, a final temperature a little below the prediction (some heat warms the mug and air). In Part B, a water temperature rise of several degrees. List two sources of error and say whether each makes your c too high or too low.
Your turn
10.100 kg of water at 90 °C is mixed with 0.100 kg of water at 30 °C. Find T_f.
Equal masses of the same substance: T_f = (90 + 30) ÷ 2 = 60 °C.
20.400 kg of water at 10 °C is mixed with 0.100 kg of water at 60 °C. Find T_f.
T_f = (0.400 × 10 + 0.100 × 60) ÷ 0.500 = (4 + 6) ÷ 0.500 = 20 °C.
3In the last question, how much heat did the hot water lose?
Q = (0.100)(4180)(20 − 60) = −1.67 × 10⁴ J (it lost 16 720 J).
4A 0.200 kg piece of iron at 100 °C is placed in 0.500 kg of water at 20 °C. Find T_f. (c(Fe) = 450)
(0.200)(450)(T_f − 100) + (0.500)(4180)(T_f − 20) = 0 → 90T_f − 9000 + 2090T_f − 41 800 = 0 → 2180T_f = 50 800 → T_f ≈ 23 °C.
5Why do we use foam cups with a lid for a calorimeter instead of a metal can?
Foam is a poor conductor (it traps air), and the lid reduces convection and evaporation, so very little heat is exchanged with the room. A metal can would conduct heat away quickly.
6A 0.100 kg metal sample at 100 °C is placed in 0.150 kg of water at 22.0 °C. T_f = 26.0 °C. Find c(metal).
Water gains (0.150)(4180)(4.0) = 2508 J. Metal: (0.100)c(74.0) = 7.40c. c = 2508 ÷ 7.40 ≈ 339 J/(kg·°C).
More practice: warm-up
1State the principle used in all mixture problems.
Heat lost by the hot substance = heat gained by the cold substance (conservation of energy).
2What is an isolated system?
A system that exchanges no energy with its surroundings.
3Two equal masses of water at 40 °C and 70 °C are mixed. Find T_f.
4In a mixture, which sample has a negative Q?
The hotter sample (it loses heat).
5Write the formula for percent error.
Percent error = |experimental − accepted| ÷ accepted × 100%.
More practice: on your own
1How much water at 15 °C must be added to 0.60 kg of water at 85 °C to make a mixture at 40 °C?
Heat lost by hot = heat gained by cold: (0.60)(45) = m(25) (c cancels). m = 27 ÷ 25 = 1.08 kg ≈ 1.1 kg.
2A student finds c(aluminium) = 820 J/(kg·°C). The accepted value is 900. Find the percent error.
|820 − 900| ÷ 900 × 100% = 80 ÷ 900 × 100% ≈ 8.9%.
3Suggest two changes to the method in Example 2 that would reduce error.
Examples: transfer the metal faster; use a lid and stir; use more insulation; start the water a few degrees below room temperature so it gains as much from the room as it loses; use a more precise thermometer; dry the metal quickly.
4A 0.250 kg copper block at 150 °C is placed in 0.400 kg of water at 18 °C. Find T_f. (c(Cu) = 390)
(0.250)(390)(T_f − 150) + (0.400)(4180)(T_f − 18) = 0 → 97.5T_f − 14 625 + 1672T_f − 30 096 = 0 → 1769.5T_f = 44 721 → T_f ≈ 25 °C.
5If heat leaks from the calorimeter to the room during an experiment to find c of a metal, will the calculated c be too high or too low? Explain.
Too low. The water's temperature rise is smaller than it should be, so the calculated heat gained is too small. Dividing a too-small heat by the metal's (correct) mass × ΔT gives too small a c.
Challenge
1A 0.120 kg aluminium calorimeter cup holds 0.300 kg of water, both at 20.0 °C. A 0.200 kg piece of iron at 95.0 °C is added. Find T_f, including the cup. (c: Al 900, Fe 450, water 4180)
Iron loses: (0.200)(450)(T_f − 95) = 90T_f − 8550. Cup gains: (0.120)(900)(T_f − 20) = 108T_f − 2160. Water: (0.300)(4180)(T_f − 20) = 1254T_f − 25 080. Sum = 0: 1452T_f = 35 790 → T_f ≈ 24.6 °C.
2Explain why the final temperature of a mixture of a hot metal and cold water is almost always much closer to the water's starting temperature.
Water has a very large specific heat capacity (often a larger mass too), so it has a much larger heat capacity (mc). The same amount of heat changes the water's temperature only a little while changing the metal's temperature a lot.
3Three samples of water are mixed: 1.0 kg at 10 °C, 2.0 kg at 40 °C and 1.0 kg at 70 °C. Find the final temperature and show that the total heat gained equals the total heat lost.
T_f = (1.0×10 + 2.0×40 + 1.0×70) ÷ 4.0 = 160 ÷ 4.0 = 40 °C. Gained by the 10 °C sample: (1.0)(4180)(30) = 125 400 J. Lost by the 70 °C sample: (1.0)(4180)(30) = 125 400 J. The 40 °C sample has ΔT = 0. Gains equal losses.
Quiz
1. In an isolated system, a hot object and a cold object reach equilibrium. Which statement is true?
2. 0.300 kg of water at 70 °C is mixed with 0.100 kg of water at 10 °C. What is T_f?
3. Why is a calorimeter made of foam cups with a lid?
4. A metal at 100 °C is placed in cool water. Water gains 2000 J. Ignoring losses, how much heat did the metal lose?
5. A student's experimental c for copper is 351 J/(kg·°C). The accepted value is 390 J/(kg·°C). What is the percent error?
6. Heat escaping to the room during a method-of-mixtures experiment usually makes the calculated specific heat capacity of the metal:
Big idea: expectation D2.9 asks students to determine specific heat capacity by mixing, and D2.10 to solve the matching problems. The key idea is conservation of energy: heat lost = heat gained (or Q_hot + Q_cold = 0).
How to help: Do Part A of the activity together; predicting the bath or tea temperature and then checking is very convincing. Ask your teen which way each source of error pushes the answer.
Common difficulties:
- Sign errors. Either write heat lost = heat gained with both sides positive, or write Q₁ + Q₂ = 0 with ΔT = T_f − T_initial for each. Do not mix the two methods.
- Using the weighted-average shortcut for two different substances (it only works for the same substance).
- Forgetting the container's heat capacity when a question gives it (Challenge 1).
- Algebra: collecting the T_f terms. Encourage writing every step.
Safety: hot tap water and boiling water burn. A grown-up should handle hot water at home; in school labs, follow the teacher's instructions for hot plates and tongs.