Ever wondered why a metal spoon gets hot faster than a wooden one in the same cup of tea? That’s calorimetry in action – and it’s the secret behind many exam questions.
Calorimetry is simply about figuring out how much heat moves from one thing to another. Think of it like tracking how water flows from one bucket to another – only the ‘water’ is heat energy.
What is Calorimetry?
Calorimetry is the branch of physics that measures the amount of heat (thermal energy) transferred during a physical change or chemical reaction. A calorimeter is the device used – imagine a insulated box that keeps heat from escaping, just like a thermos bottle keeps your coffee warm.
Key Formulae for Heat Exchange
The core equation you’ll use over and over is:
Q = m × c × Δt
- Q = heat transferred (in joules, J)
- m = mass of the substance (in kilograms, kg)
- c = specific heat capacity – the amount of heat needed to raise 1 kg of a material by 1°C (J kg⁻¹ °C⁻¹). Think of it as the ‘thickness’ of a material’s heat‑storage blanket.
- Δt = change in temperature (final temperature minus initial temperature, in °C)
When two bodies exchange heat inside a calorimeter, the heat lost by the hotter body equals the heat gained by the cooler one (assuming no heat loss to the surroundings). In symbols:
m₁c₁Δt₁ = m₂c₂Δt₂
Step‑by‑Step Method to Solve a Calorimetry Problem
Follow these four tidy steps and you’ll never get stuck.
1. Read the question carefully
Mark what you know: masses, initial temperatures, material types, and what the question asks for.
2. Identify hot and cold bodies
The body with the higher initial temperature will lose heat (its Δt will be negative), while the cooler one gains heat (Δt positive).
3. Write the heat‑exchange equation
Use m₁c₁Δt₁ + m₂c₂Δt₂ = 0 or rearrange to m₁c₁Δt₁ = -m₂c₂Δt₂. The minus sign simply reminds you that one Δt is negative.
4. Plug in numbers and solve
Insert the given masses, specific heats (look them up in the textbook table), and temperature differences. Solve for the unknown temperature or mass.
Worked Example 1 – Finding the Final Temperature
Question: A 150 g piece of copper (c = 0.385 J g⁻¹ °C⁻¹) at 80°C is dropped into 200 g of water (c = 4.18 J g⁻¹ °C⁻¹) initially at 25°C. Assuming no heat loss, find the equilibrium temperature.
Solution:
- Identify hot body: copper (80°C). Cold body: water (25°C).
- Let the final temperature be Tₓ.
- Write heat balance:
m_cu·c_cu·(Tₓ‑80) + m_w·c_w·(Tₓ‑25) = 0 - Insert numbers (convert grams to kilograms or keep grams consistently – here we stay in grams):
150×0.385×(Tₓ‑80) + 200×4.18×(Tₓ‑25) = 0 - Expand and collect Tₓ terms:
57.75(Tₓ‑80) + 836(Tₓ‑25) = 0
57.75Tₓ‑4620 + 836Tₓ‑20900 = 0
(57.75+836)Tₓ = 4620+20900
893.75Tₓ = 25520 - Divide: Tₓ ≈ 28.6°C. So the copper cools down a lot, while the water warms just a little.
Worked Example 2 – Determining Unknown Mass
Question: 100 g of aluminium (c = 0.900 J g⁻¹ °C⁻¹) at 70°C is placed in a calorimeter containing 250 g of oil (c = 2.00 J g⁻¹ °C⁻¹) at 30°C. The final temperature recorded is 35°C. Find the mass of oil that actually participated in heat exchange (the rest is assumed to be insulated).
Solution:
- Hot body: aluminium. Cold body: oil.
- Heat lost by aluminium = heat gained by oil.
m_al·c_al·(T_final‑T_initial_al) = m_o·c_o·(T_final‑T_initial_o) - Plug known values (Δt for aluminium = 35‑70 = -35°C, for oil = 35‑30 = 5°C):
100×0.900×(-35) = m_o×2.00×5 - Calculate left side: 100×0.900 = 90; 90×(-35) = -3150 J.
So, -3150 = 10 m_o - Solve: m_o = 315 g.
- Thus, 315 g of oil absorbed the heat; any extra oil beyond this amount stayed thermally isolated.
Quick Summary Table
| Step | What to Do | Key Point |
|---|---|---|
| 1 | Read the problem, note all given data | Units must match (g with J g⁻¹ °C⁻¹) |
| 2 | Identify which object loses heat and which gains | Hot → negative Δt, Cold → positive Δt |
| 3 | Write the heat‑exchange equation | m₁c₁Δt₁ = -m₂c₂Δt₂ |
| 4 | Plug numbers, solve for the unknown | Check sign and keep significant figures |
Common Mistakes to Avoid
- Forgetting the minus sign when one body loses heat.
- Mixing up specific heat values – aluminium is 0.900 J g⁻¹ °C⁻¹, copper is 0.385 J g⁻¹ °C⁻¹.
- Leaving the calorimeter’s own heat capacity out. In ICSE questions it’s usually given; treat it as another ‘body’ in the equation.
- Not converting grams to kilograms when the formula expects kg. Stay consistent.
📝 Likely Exam Questions
- Question: A 50 g metal block at 120°C is placed in 100 g of water at 20°C. The metal’s specific heat capacity is 0.200 J g⁻¹ °C⁻¹. Find the final temperature assuming no heat loss. Answer: Use heat balance: 50×0.200×(T‑120) + 100×4.18×(T‑20)=0 → T≈31°C.
- Question: In a calorimeter, 80 g of ice at 0°C melts completely and the resulting water reaches 25°C. If the calorimeter’s heat capacity is 150 J °C⁻¹, calculate the heat absorbed by the ice. (Latent heat of fusion of ice = 334 J g⁻¹.) Answer: Heat to melt ice: 80×334 = 26,720 J. Heat to raise water from 0°C to 25°C: 80×4.18×25 = 8,380 J. Total = 35,100 J. Calorimeter absorbs 150×25 = 3,750 J, so heat from ice = 35,100 J – 3,750 J = 31,350 J.
- Question: A 200 g piece of brass (c = 0.380 J g⁻¹ °C⁻¹) at 90°C is dropped into 300 g of oil (c = 2.00 J g⁻¹ °C⁻¹) initially at 25°C. If the final temperature is 30°C, find the heat lost by brass. Answer: Δt_brass = 30‑90 = -60°C. Heat lost = 200×0.380×(-60) = -4,560 J (magnitude 4,560 J).
- Question: Explain why a calorimeter must be insulated during an experiment. Answer: Insulation prevents heat exchange with the surroundings, ensuring that the measured temperature change reflects only the exchange between the substances inside, which is the assumption behind the heat‑balance equation.