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Exam level2 past questions

Thermal properties of matter

Deals with heat transfer, thermal conductivity, calorimetry, specific heat, and thermal expansion of solids.

Why this shows up in the exam

You need to solve NEET problems involving heat flow, temperature changes, and expansion of materials.

How NEET tests this

Numerical · 3 QsApplication

Learn the idea

Thermal properties link heat exchanged to mass, specific heat and temperature change; the breakthrough is that in a calorimeter the heat lost by the hot body exactly equals the heat gained by the water plus the calorimeter’s water‑equivalent.

🧠 Memory hook: Think of the calorimeter as a twin of water – its heat capacity is just extra water mass.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Specific heat c = heat absorbed ÷ (mass × ΔT) (J g⁻¹ K⁻¹)
  • Water equivalent W of a calorimeter acts like an equal mass of water with c = 4.2 J g⁻¹ K⁻¹
  • In an isolated calorimetry set‑up ΣQ_gained = ΣQ_lost
  • Thermal conductivity K governs heat flow Q̇ = K·A·ΔT⁄L
  • Linear expansion ΔL = α·L₀·ΔT for solids

How to approach it

  1. 1Write the heat‑balance equation: heat lost by hot object = heat gained by water + calorimeter
  2. 2Replace each heat term by m·c·ΔT, using c_water = 4.2 J g⁻¹ K⁻¹ and W as an equivalent mass of water
  3. 3Solve the equation for the unknown specific heat
  4. 4Convert all masses to the same unit and keep ΔT positive

Worked example — watch it click

In an experiment on the specific heat of a metal, a 0.2 kg block of the metal at 150° C is dropped in a copper calorimeter (of water equivalent of 0.025 kg) containing 150 cm³ of water at 27° C. The final temperature is 40° C, the specific heat of metal will be

  • A)0.4 Jg⁻¹ K⁻¹
  • ✅0.43 Jg⁻¹ K⁻¹
  • C)0.54 Jg⁻¹ K⁻¹
  • D)0.61 Jg⁻¹ K⁻¹

The concept behind this problem

The problem forces you to set up a calorimetric heat balance that includes the water‑equivalent, a direct test of the heat‑loss = heat‑gain principle.

Step by step

  1. 1Heat lost by metal = Heat gained by water + calorimeter. m_metal × c_metal × ΔT_metal = (m_water × c_water + W) × ΔT_water.
  2. 2Given: m_metal = 0.2 kg = 200 g, T_initial_metal = 150°C, m_water = 150 cm³ = 150 g, T_initial_water = 27°C, T_final = 40°C, W = 0.025 kg = 25 g (water equivalent). 200 × c × (150-40) = (150 × 4.2 + 25 × 4.2) × (40-27). 200 × c × 110 = 175 × 4.2 × 13 = 9555. c = 9555/(200 × 110) = 9555/22000 = 0.434 J/g/K ≈ 0.43 J/g/K.

Watch out

Students often take the water’s ΔT as 27‑40 °C (negative) instead of 40‑27 °C, giving a wrong sign in the heat‑gain term.

Common slip-ups that cost marks

  • •Using opposite sign for a temperature difference
  • •Mixing kg and g without conversion
  • •Assuming the calorimeter contributes no heat capacity

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 2

In an experiment on the specific heat of a metal, a 0.2 kg block of the metal at 150° C is dropped in a copper calorimeter (of water equivalent of 0.025 kg) containing 150 cm³ of water at 27° C. The final temperature is 40° C, the specific heat of metal will be

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