In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
(i) Specific heat of metal = 432.7 J kg⁻¹ K⁻¹ (or 0.43 kJ kg⁻¹ K⁻¹)
(ii) The calculated value is smaller than the actual specific heat of the metal.
Explanation
This numerical problem applies the principle of calorimetry from the chapter: heat lost by hot body equals heat gained by cold body. The water equivalent concept simplifies calorimeter heat capacity calculation. The second part tests understanding of experimental errors - heat losses cause the calculated specific heat to be underestimated.
Solution Steps
Step 1: Calculate mass of water. Volume = 150 cm³, so mass = 150 g = 0.150 kg (since density of water = 1000 kg/m³)
Step 2: Apply principle of calorimetry: Heat lost by metal = Heat gained by water + Heat gained by calorimeter
Step 3: Calculate temperature changes. For metal: ΔT₁ = 150 - 40 = 110 °C. For water and calorimeter: ΔT₂ = 40 - 27 = 13 °C
Step 4: Set up equation: m₁ × s × ΔT₁ = (m₂ + W) × s_w × ΔT₂, where W = water equivalent = 0.025 kg, s_w = ³ J kg⁻¹ K⁻¹
Step 5: Substitute values: 0.20 × s × 110 = (0.150 + 0.025) × ³ × 13
Step 6: Solve: 22s = × 13 = 9518.5, therefore s = 432.7 J kg⁻¹ K⁻¹
Step 7: For heat loss analysis: If heat escapes to surroundings, actual heat lost by metal > measured heat gained. Since s ∝ (heat gained)/(m × ΔT), the calculated s is smaller than actual value.