Question

A copper vessel contains 200 gram of water at 24°C. 112 gram of water at 42° C is added, the resultant temperature of water is 30°C. Calculate the thermal capacity of the calorimeter.​

Answer 1

The heat capacity depends is the heat absorbed by the calorimeter as

the temperature changes.

Response:

• The thermal capacity of the calorimeter is 100.416 J·K⁻¹

What the thermal capacity and how can it be calculated?

Mass of the vessel. m₁ = 200 gram

Temperature of the water, T₁  = 24°C

Mass of the warm water, m₂ = 112 grams

Temperature of the warm water, T₂ = 42°C

Resultant temperature of the water, T₃ = 30°C

The heat capacity of the water = 4.184 J·Kg⁻¹·K⁻¹

• Heat lost by the warm water = Heat gained by the cold water + The heat gained by the calorimeter

Let C represent the thermal capacity of the calorimeter

We have;

Change in heat, ΔH = m·c·ΔT

Therefore;

112 × 4.184 × (42 - 30) = 200 × 4.184 × (30 - 24) + C×(30 - 24)

5623.296 = 5020.8 + 6·C

6·C = 5623.296 - 5020.8 = 602.496

$The \ heat \ capacity \ of \ the \ calorimeter, \ C = \dfrac{602.496}{6} = \mathbf{ 100.416}$

• The thermal capacity of the calorimeter, C = 100.416 J·K⁻¹

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