A sample of NaOH(s) was added to water in a calorimeter. The temperature was monitored as the NaOH dissolved to give the data below. Determine the heat released during the solution process. (Assume the solution specific heat is 4.18 J•g–1•K–1)

Mass of water 100.00 g
Mass of NaOH(s) 10.00 g
Initial Temperature of water 24.0 °C
Final Temperature of solution 48.2 °C

(A) 1.01 * 103Joules
(B) 2.66 * 103Joules
(C) 1.01 * 104Joules
(D) 1.11* 104Joules

According to the law of conservation of energy, the sum of the heat released by the solution of NaOH and the heat absorbed by the aqueous dissolution is zero.

Qsol(NaOH) + Qaq,dis = 0

Qsol(NaOH) = - Qaq,dis [1]

The heat absorbed by the aqueous dissolution can be calculated using the following expression.

Qaq,dis = c × m × ΔT

where,

c: specific heat of the solution (4.18 J.g⁻¹.K⁻¹ = 4.18 J.g⁻¹.°C⁻¹)

m: mass of the aqueous dissolution (100.00 g + 10.00 g = 110.00 g)

ΔT: change in the temperature (48.2°C - 24.0°C = 24.2°C)

Qaq,dis = 4.18 J.g⁻¹.°C⁻¹ × 110.00 g × 24.2°C = 1.11 × 10⁴ J

From [1],

Qsol(NaOH) = -1.11 × 10⁴ J

The amount of heat released during the solution process is 1.11 × 10⁴ J.