Calorimetry & Specific Heat — A/L Physics

How do I practise Calorimetry & Specific Heat?

1 question with a worked solution

1. A 0.10 kg block of copper at 100 °C is dropped into 0.20 kg of water at 20 °C in a copper calorimeter of mass 0.05 kg. Specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and copper is 400 J kg⁻¹ K⁻¹. Assuming no heat loss to the surroundings, what is the final equilibrium temperature?
   1. A) 21.8 °C
   2. B) 23.6 °C
   3. C) 28.4 °C
   4. D) 25.0 °C
   Check my answer

   Correct answer: B) 23.6 °C

   Heat lost by hot copper = Heat gained by water and calorimeter. Heat capacity of copper block = 0.10 × 400 = 40 J K⁻¹. Heat capacity of water + calorimeter = (0.20 × 4200) + (0.05 × 400) = 840 + 20 = 860 J K⁻¹. Equation: 40 × (100 - θ) = 860 × (θ - 20) => 4000 - 40θ = 860θ - 17200 => 900θ = 21200 => θ = 21200 / 900 ≈ 23.56 °C (approx 23.6 °C).

   Show worked solution

   Step 1: m_c c_c (100 - θ) = (m_w c_w + m_cal c_c)(θ - 20). Step 2: 40(100 - θ) = 860(θ - 20) => 4000 - 40θ = 860θ - 17200. Step 3: 900θ = 21200 => θ ≈ 23.56 °C. Answer: 23.6 °C.
   Modelled on: the A/L Physics examination pattern

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Where does Calorimetry & Specific Heat fit in the syllabus?

Unit 4.4 · Year 1 · Term 2