A 6.25 kg block of an unknown substance with a temperature of 104.0°C is placed in a calorimeter con

Question

A 6.25 kg block of an unknown substance with a temperature of 104.0°C is placed in a calorimeter containing 16.3 kg of water at 34.6°C. The system reaches an equilibrium temperature of 37.0°C. What is the specific heat capacity of the unknown substance if the heat capacity of the calorimeter is neglected?

[Specific heat capacity of water, c = 4200 Jkg⁻¹K⁻¹]

Answer 1

A 6.25 kg block of an unknown substance with a temperature of 104.0°C is placed in a calorimeter containing 16.3 kg of water at 34.6°C. The system reaches an equilibrium temperature of 37.0°C. What is the specific heat capacity of the unknown substance if the heat capacity of the calorimeter is neglected?

[Specific heat capacity of water, c = 4200 Jkg⁻¹K⁻¹]

4.2 x 10² Jkg⁻¹K⁻¹
3.9 x 10² Jkg⁻¹K⁻¹ ✓
1.4 x 10² Jkg⁻¹K⁻¹
4.5 x 10² Jkg⁻¹K⁻¹

Explanation

Heat lost by the block = heat gained by the water.

Water gains 16.3 × 4200 × (37.0 − 34.6) = 16.3 × 4200 × 2.4 = 164,304 J.

The block loses 6.25 × c × (104.0 − 37.0) = 418.75c. So c = 164,304 ÷ 418.75 = 392 ≈ 3.9 × 10² J kg⁻¹ K⁻¹.

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