(a) At constant-volume, Q = 66.7kJ
(b) At isobaric condition, Q = 92.7kJ
Explanation:
The specific heats of oxygen at the average temperature of (20+120)/2 = 70°C = 343 K are
Cp = 0.927 kJ/kg.K
Cv = 0.667 kJ/kgK
We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as
ΔE (system) = E(in) - E(out)
and
Q(in) = ΔH = mCp x (T₂ - T₁)
since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium
process. Substituting for both cases,
(a)
Qin,V =const = mCv (T2 - T1 )
= (1kg)(0.667 kJ/kg.K)(120 - 20)K
= 66.7kJ
(b)
Qin,P=const = mCp (T2 - T1 )
= (1kg)(0.927 kJ/kg.K)(120 - 20)K
= 92.7kJ
