Correct Answer - Option 3 : 2.03 kJ/kg°C
Concept:
Specific enthalpy of superheated steam is given by:
ht2 - ht1 = Cps(t2 - t1)
where ht2 is specific enthalpy at temperature t2 and ht1 is specific enthalpy at temperature t1
Calculation:
Given:
ht2 = 3278.9 kJ/kg, ht1 = 3075.5 kJ/kg, t2 = 400°C, t1 = 300°C, Cps = mean specific heat
We know that,
ht2 - ht1 = Cps(t2 - t1)
3278.9 - 3075.5 = Cps × (400 - 300)
203.4 = Cps × 100
Cps = \(\frac{203.4}{100}=2.03\) kJ/kg°C