Answer: Tfinal = 7.1°C
Explanation:
heat released or absorbed = mass × specific heat capacity × change in temperature
q = m × cg × ΔT (eqn 1)
Note: ΔT = (Tfinal - Tinitial)
(q = ? ΔHsoln=25.7kJ/mole = 25700J/mole; mass of solution, m = 100 + 35g = 135g; cg = specific heat capacity of water = 4.18 J°C-1g-1; ΔT = ? masss of solute, NH4NO3 = 35g, molar mass of solute, NH4NO3 = 80g)
molar enthalpy of solution, ΔHsoln
<em> = heat absorbed or released ÷ moles of solute, </em><em>n</em>
ΔHsoln = q ÷ n
q = ΔHsoln × n
<em>moles solute</em>, n = mass solute (g) ÷ molar mass solute (g mol-1)
moles of solute, n = 35g/80g/mol = 0.4375 moles
q = 25700J/mol × 0.4375 mol = 11243.75J
From equation 1 above, ΔT = q / (m × cg) = 11243.75J / (135 × 4.18 J°C-1g-1) = 19.9°C
Since the reaction is endothermic, Tinitial > Tfinal, therefore, Tfinal = Tinitial - ΔT = 27 - 19.9 = 7.1°C
