Answer to Question #122897 in Molecular Physics | Thermodynamics for Favour

Question

The specific heat at constant pressure of 1kg fluid undergoing a MF fluid constant pressure process is given by
Cp(2.5+40÷T+20)kg/kg⁰C
Where T is in ⁰C.
The pressure during the process is maintained at 2 bar and volume changes from 1m³ to 1.8m³ and temperature changes from 50⁰C to 450⁰C. Determine
(i)Heat added
(ii)Work done
(iii)Change in internal Energy
(iv)Change in Enthalpy

Answer 1

(i) Heat added 

$Q = m C_Pdt$

$Q=m \times Cp(2.5+40÷T+20)\times dt$

$Q= 1 \times 4.187(2.5+40÷400+20)\times 400=37.85kJ$

(ii) Work done  

$W = p (V_2-V_1)=200000(1.8-1)=160 kJ$

(iii) Change in internal Energy:

$U = W - Q =160-37.85 =122.15 kJ$

(iv) Change in Enthalpy:

$dH = dQ =Q = 37.85 kJ$