Answer:
(a)
dQ = mdq
dq = [tex]C_p[/tex]dT
[tex]q = \int\limits^{T_2}_{T_1} {C_p} \, dT[/tex] = [tex]C_p[/tex] (T₂ - T₁)
From the above equations, the underlying assumption is that [tex]C_p[/tex] remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let [tex]C_{cal}[/tex] be heat constant of calorimeter
Q₂ = [tex]C_{cal}[/tex] ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m [tex]C_p[/tex] ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m [tex]C_p[/tex]' ΔT
[tex]C_p[/tex] = (16.73 - 6.14) / (15.085 x 3.10)
[tex]C_p[/tex] = 0.22646 KJ mol⁻¹ k⁻¹