Ever wondered why a hot cup of tea cools down while the room stays almost the same? That's calorimetry in action – the science of heat exchange.
💡 In Simple Words: Calorimetry is all about tracking how much heat moves from a hotter object to a cooler one. When you know the masses, temperatures, and material properties, you can predict the final temperature.
What is Calorimetry?
Calorimetry (pronounced ka‑lo‑ri‑me‑tree) is the study of heat transfer between objects. In physics class we use a special container called a calorimeter to keep the exchange tidy, but the ideas work for any mixing of hot and cold things.
Key Terms You Need
- Specific heat (c): the amount of heat needed to raise 1 kg of a substance by 1 °C. Think of it as the "heat‑holding capacity" of a material.
- Heat capacity (C): similar to specific heat but for a whole object, not per kilogram. It’s just mass × specific heat.
- Calorimeter: a vessel (often insulated) where we let two substances exchange heat while we measure temperatures.
- Heat exchange (Q): the quantity of heat that moves. Positive Q means heat gained, negative means heat lost.
The Core Formula
For any material, the heat exchanged is given by
Q = m c ΔT
where m is mass, c is specific heat, and ΔT (delta‑T) is the change in temperature (final minus initial). If you have more than one substance, the heat lost by the hot side equals the heat gained by the cold side (ignoring losses to the surroundings).
Steps to Solve Heat Exchange Problems
Follow these four simple steps, and you’ll never get stuck.
Worked Example 1: Mixing Two Water Samples
Problem: 200 g of water at 80 °C is mixed with 300 g of water at 20 °C in a perfect calorimeter. Find the final temperature.
Solution:
- Identify data: mass₁ = 0.200 kg, T₁i = 80 °C; mass₂ = 0.300 kg, T₂i = 20 °C; specific heat of water c = 4186 J kg⁻¹ °C⁻¹.
- Write Q for each part: Q₁ = m₁c(T_f‑T₁i), Q₂ = m₂c(T_f‑T₂i).
- Set heat lost = heat gained: because the hotter water cools, Q₁ will be negative; the colder water warms, Q₂ positive. So
m₁c(T_f‑80) + m₂c(T_f‑20) = 0. - Simplify (c cancels): 0.200(T_f‑80) + 0.300(T_f‑20) = 0 → 0.200T_f‑16 + 0.300T_f‑6 = 0 → 0.500T_f‑22 = 0 → T_f = 44 °C.
- Check: hotter water loses heat (44‑80 = –36 °C), colder gains (44‑20 = +24 °C). The signs line up, so 44 °C is correct.
Worked Example 2: Metal Block in Water
Problem: A 150 g copper block (c = 385 J kg⁻¹ °C⁻¹) at 100 °C is dropped into 250 g of water at 25 °C (c = 4186 J kg⁻¹ °C⁻¹). Find the equilibrium temperature.
Solution:
- Convert masses to kilograms: m_Cu = 0.150 kg, m_w = 0.250 kg.
- Write heat balance: m_Cu c_Cu(T_f‑100) + m_w c_w(T_f‑25) = 0.
- Plug numbers: 0.150·385(T_f‑100) + 0.250·4186(T_f‑25) = 0 → 57.75(T_f‑100) + 1046.5(T_f‑25) = 0.
- Expand: 57.75T_f‑5775 + 1046.5T_f‑26162.5 = 0 → 1104.25T_f‑31937.5 = 0 → T_f = 28.9 °C (≈ 29 °C).
- Interpretation: copper cools dramatically, water warms a little – the final temperature is just above the initial water temperature.
Quick Comparison Table
| Situation | What to Write | Key Tip |
|---|---|---|
| Two liquids mixing | m₁c(T_f‑T₁i) + m₂c(T_f‑T₂i) = 0 | c is same if both are water. |
| Solid in liquid | m_s c_s(T_f‑T_s_i) + m_l c_l(T_f‑T_l_i) = 0 | Keep track of sign – hot side negative. |
| Calorimeter with known heat capacity C_cal | Q_lost + Q_cal = 0 | Use C_cal directly, no mass needed. |
Common Mistakes to Avoid
- Forgetting to convert grams to kilograms – the formula expects kg.
- Mixing up ΔT direction. Write final minus initial, not the other way around.
- Leaving out the calorimeter’s own heat capacity when it’s given.
- Assuming heat is lost to the environment; in exam problems we treat the system as isolated unless told otherwise.
📝 Likely Exam Questions
- Question: 100 g of iron (c = 450 J kg⁻¹ °C⁻¹) at 150 °C is placed in 200 g of water at 20 °C. Find the final temperature.
- Model Answer: Set up 0.100·450(T_f‑150) + 0.200·4186(T_f‑20) = 0. Solve → T_f ≈ 23 °C.
- Question: A calorimeter of heat capacity 150 J °C⁻¹ contains 250 g of water at 30 °C. 50 g of steam at 100 °C condenses inside. Find the final temperature.
- Model Answer: Heat gained = heat lost: m_steam·L_v + m_steam·c_water(T_f‑100) = C_cal(T_f‑30) + m_water·c_water(T_f‑30). Solve → T_f ≈ 44 °C.
- Question: Explain why the heat lost by a hot object equals the heat gained by a cooler one in a closed calorimeter.
- Model Answer: Energy cannot be created or destroyed (law of conservation). In an insulated system the only energy transfer is heat, so what one part loses, another must gain.