Question

A 6.40 g sample of solid KCl was dissolved in 42.0 g of water. The initial temperature of the water was 20.60°C. After the compound dissolved, the temperature of the water was 11.30°C. Assume the heat was completely absorbed from the water and no heat was absorbed by the reaction container or the surroundings. Calculate the heat absorbed by the process. The specific heat of water is 4.184 J/g·°C.

Answer 1

The heat absorbed by the process is ‭-1,634.27‬ J

Step-by-step explanation:

Here, we have

Mass of KCl = 6.4 g

Mass of water = 42 g

Initial temperature of water = 20.60°C

Final temperature of the water = 11.30°C

Specific heat capacity of water = 4.184 J/g·°C

Based on the principle of conservation of energy, which states that energy can neither be created nor destroyed, but changes from one form to another, we have

The heat absorbed in the process is equal to the heat lost by the water present in the dissolution

Heat lost by water = ΔH = 42 × 4.184 × (11.3 - 20.6) = ‭-1,634.2704‬ J

Heat lost by water = 1.63 kJ

∴ The heat absorbed by the process = ‭-1,634.2704‬ J