Determine the specific heat capacity of different materials. Supply a known amount of heat and measure the temperature rise, solve Q = mcΔT for any of its quantities, and find the specific heat of an unknown metal by calorimetry — heat lost by the hot metal equals heat gained by the water. In each section, calculate the answer yourself first, then run the experiment and compare.
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Theory — Specific Heat Capacity
Heat, Mass, and Temperature Change
The specific heat capacity (c) of a material is the heat needed to raise the temperature of 1 gram of it by 1 °C. Water has a high specific heat (4.18 J/g°C), while metals have low ones, so metals heat up and cool down quickly.
Heat Equation
Q = m c ΔT
Q = heat (J) · m = mass (g) · c = specific heat (J/g°C) · ΔT = temperature change (°C)
c = Q / (m ΔT) · m = Q / (c ΔT) · ΔT = Q / (m c)
Determining Specific Heat by Calorimetry
To find the specific heat of an unknown metal, heat it, then drop it into cooler water in an insulated cup. The heat lost by the metal equals the heat gained by the water (energy is conserved), and both reach the same final temperature.
Calorimetry — Energy Balance
heat lost by metal = heat gained by water
m(metal) × c(metal) × (T(hot) − T(final)) = m(water) × c(water) × (T(final) − T(cold))
Heat to raise 1 g by 1 °C. Low for metals, high for water.
Q = mcΔT
Links heat, mass, specific heat, and temperature change.
Calorimetry
Heat lost = heat gained; used to find an unknown specific heat.
Material
Specific heat (J/g°C)
Water
4.18
Aluminium
0.90
Iron
0.45
Copper
0.385
Lead
0.13
Apparatus
The equipment a real specific-heat experiment uses to supply a known amount of heat and measure the temperature change. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.
Coffee-cup calorimeter
An insulated cup so the heat lost by the metal goes into the water, not the surroundings.
Thermometer
Measures the temperature change ΔT and the final temperature.
Analytical balance
Weighs the sample and the water for Q = mcΔT.
Bunsen burner
Heats the metal sample to its starting temperature.
Beaker
Holds the water that receives the heat.
Stopwatch
Times the heating so the heat supplied Q is known.
Instructions — Running the Virtual Experiment
The simulation has three sections. In each one, calculate the answer yourself first, then run the experiment to check it, and record both values with screenshots. Repeat with at least two different materials so you can compare specific heats.
Part 1 — Heat a Substance (Heat a Substance tab)
1
Open Simulation → Heat a Substance. Choose a material and run it. Read the mass, the heat supplied (Q), and the temperature change (ΔT). Calculate the specific heat yourself with c = Q/(mΔT), type it in, then check. Identify the material from its specific heat.
2
Repeat for at least two different materials and compare their specific heats.
Part 2 — The Heat Equation Q = mcΔT (Q = mcΔT tab)
1
Open Q = mcΔT. Each trial gives you three of the four quantities and asks for the fourth (Q, c, ΔT, or m). Rearrange the equation and calculate the unknown yourself, type it in, then check.
Part 3 — Calorimetry (Calorimetry tab)
1
Open Calorimetry. Choose an unknown hot metal sample, drop it into the water, and read the final temperature. Calculate the metal's specific heat yourself from heat lost = heat gained, type it in, then check and identify the metal.
Simulation — Specific Heat Capacity
Specific Heat Virtual LabHeat · Q = mcΔT · Calorimetry
Specific heat from heating
Q = m c ΔT → c = Q ÷ (m ΔT)
Then identify the material
Compare your c with the reference table to name the material.
Drop the metal in, work out c, type it, then check.
Team Questions
Question 1. How much heat is needed to raise 50 g of aluminium (c = 0.90 J/g°C) by 30 °C? (Q = mcΔT; type in J)
Question 2. A 100 g sample absorbs 1925 J and rises 50 °C. What is its specific heat? (c = Q/(mΔT); type to 3 decimals)
Question 3. From Q2, which material is it (compare to the table)? (Type the name)
Question 4. 3600 J is added to 200 g of iron (c = 0.45 J/g°C). What is the temperature change? (ΔT = Q/(mc); type in °C)
Question 5. Why does water heat up more slowly than a metal of the same mass? (Answer in a phrase)
Question 6. In calorimetry, what is the relationship between the heat lost by the hot metal and the heat gained by the water? (Answer in a phrase)
Question 7 — Challenge. A 100 g metal at 200 °C is dropped into 150 g of water at 20 °C; the final temperature is 30.4 °C. Find the specific heat of the metal. (c(water) = 4.18; type to 3 decimals)
Example Lab Report
Sample report demonstrating the expected format. Include your data tables for at least two materials, the full calculations, and a comparison of the specific heats.
To determine the specific heat capacity of different materials using Q = mcΔT, and to find the specific heat of an unknown metal by calorimetry.
Theory
The heat needed to change a material's temperature is Q = mcΔT. In calorimetry, the heat lost by a hot object equals the heat gained by the cooler water, because energy is conserved.
Heating example: c = 900/(50×20) = 0.90 J/g°C → aluminium.
Calorimetry example: 100 g metal at 200 °C into 150 g water at 20 °C, final 30.4 °C. Heat gained by water = 150 × 4.18 × (30.4 − 20) = 6521 J. c(metal) = 6521 / [100 × (200 − 30.4)] = 0.385 J/g°C → copper.
Observations & Analysis
Metals had much lower specific heats than water, so they changed temperature easily. The calorimetry result matched the reference value for copper closely, with small differences due to rounding the final temperature and the assumption of a perfectly insulated cup.
Conclusion
The specific heat capacities were determined using Q = mcΔT and confirmed by calorimetry. Water's high specific heat (4.18 J/g°C) explains why it resists temperature change compared with metals.
Practice Questions
Show all work. Use Q = mcΔT and its rearrangements, and for calorimetry, heat lost by metal = heat gained by water. Specific heats (J/g°C): water 4.18, Al 0.90, Fe 0.45, Cu 0.385, Pb 0.13.
Question 1
How much heat raises 250 g of water by 15 °C?
Hint: Q = 250 × 4.18 × 15 = 15 675 J.
Question 2
A 75 g metal absorbs 1012.5 J and rises 30 °C. Find its specific heat and identify it.
Hint: c = 1012.5/(75×30) = 0.45 J/g°C → iron.
Question 3
How much will 120 g of copper (c = 0.385) rise if it absorbs 924 J?
Hint: ΔT = 924/(120×0.385) = 20 °C.
Question 4
What mass of aluminium (c = 0.90) absorbs 1350 J while rising 30 °C?
Hint: m = 1350/(0.90×30) = 50 g.
Question 5
An 80 g metal at 150 °C is dropped into 120 g of water at 22 °C; the final temperature is 38.1 °C. Find the metal's specific heat and identify it.
