Respuesta :
Answer:
The final temperature of the NaOH solution: T₂ = 37 °C
Explanation:
Given: The enthalpy of NaOH solution: [tex]\Delta H_{sol}[/tex] = – 44.4 kJ/mol,
The specific heat capacity of solution: c = 4.184 J/g.K
Given mass of solute (NaOH): w = 13.9 g, Molar mass of NaOH: m = 40 g/mol
Mass of the solvent: W = 250 g
Initial temperature of the solution: T₁ = 23 °C = 23 + 273 = 296 K (∵ 0°C = 273 K)
Final temperature of the solution: T₂ = ?
The number of moles of NaOH = [tex]\frac{given\: mass}{molar\: mass} = \frac{13.9\, g}{40\, g/mol} = 0.3475 mol[/tex]
So the amount of heat released (ΔH) by dissolution of 13.9 g NaOH:
[tex]\Delta H = \Delta H_{sol} \times n[/tex]
[tex]\Rightarrow \Delta H = (- 44.4 kJ/mol) \times (0.3475 mol) = 15.429 kJ[/tex]
[tex]\Rightarrow \Delta H = 15.429 kJ = 15429 J[/tex]
(∵ 1 kJ = 1000J )
Since the specific heat capacity is given by the equation:
[tex]c = \frac{\Delta H}{M\times \Delta T}[/tex]
Here, M is the total mass of the solution = w + W = 13.9 g + 250 g =263.9 g
Thus the increase in the temperature (ΔT) can be calculated as:
[tex]\Delta T = \frac{\Delta H}{M\times c}[/tex]
[tex]\Rightarrow \Delta T = \frac{15429 J}{263.9g \times 4.184J/g.K} = 13.973[/tex]
[tex]\Rightarrow \Delta T = T_{2} - T_{1} = 13.973[/tex]
[tex]\Rightarrow \Delta T = T_{2} - 296K = 13.973[/tex]
[tex]\Rightarrow T_{2} = 296K + 13.97 = 309.97K = 309.97 - 273^{\circ }C = 36.97^{\circ }C[/tex]
(∵ 0°C = 273 K)
[tex]\Rightarrow T_{2} \approx 37^{\circ }C[/tex]
Therefore, the final temperature of the NaOH solution: T₂ = 37 °C
