Question

When 600 mL of 0.2 M HNO3 is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ × 10-2 °C.
(Enthalpy of neutralisation = 57 kJ mol-1 and Specific heat of water = 4.2 JK-1 g-1)
(Neglect heat capacity of flask)

Answer 1

$\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$
m moles 120 40 – 80 – 40

Heat liberated from reaction
= 40 × 10–3× 57 × 103 J …(1)

Heat gained by solution = $mC\Delta T$
Mass of solution $m = V × d = 1000 × 1 = 1000 g$
Heat gained by solution $= 1000 × 4.2 × ΔT …(2)$
From eq (1) and eq (2)
Heat liberated = Heat gained
$40 × 10–3 × 57 × 103 = 1000 × 4.2 × ΔT$
$ΔT = 54 ×$ 10–2 $°C$
$ΔT = 54°C$ (Rounded off to the nearest integer)
So, the answer is 54°C.