Question

In a variable volume container, 35 kg of water is heated from 75 °C to 250 °C at a constant pressure of 5 bar. a. [4 pts] How much heat was added to the container? b. [8 pts] If the same amount of heat is instead added to 35 kg of water at 75 °C in a rigid container at a constant pressure of 5 bar, what would be the final temperature?

Answer 1

Solution 35 Kg Ti = 75°C Ta = 250°C P =P2 - P 5 bar => For Variable Noleme System Q = AU + PAV

g mcv (T2-T,) & P (V₂ -V.) by ideal gas V₂ = mRitz 4 V=MRTI An o mcv (Tz -T.) + P P (MRTz - MRT meu (T₂-T,) + mR (Tz -T1) = (T2-T.) (mcutm L or m Lp (T2-T) - as Cp-cv=R? {p for water 4.18 KT Kgoc الم Q = 35 kg x 4.18 kg Kg °C (250-75 ) °C 25602.5 KJ Hence, the heat added to container is 25602.5 kJ

=> for Rigid Volume ( X = va) {To= 75°} Δν + ΡΔν us Vi = V2 g AU mcv(Tz-T.) mer( Tz -T.) we have 25602.5 kg (v for water 2.98 K5 Kg °C 25602.5 kJ 35 Kg x 2.98 kJ (T2-75) °c Kyc Tz 25602.5 kJ ol + 75 °C 35 kg x 2.98 kJ T 320-46 °C Hence, The Final temperature is is 320.46 °C