Answer to Question #215466 in Mechanical Engineering for Aron

Question

Assume that a 2 kg of a certain gas with R=0.4305 kJ/kg-K undergoes a process that results in these changes: ΔU=1458.01 kJ, ΔH=2189.13 kJ. What is the temperature change?

Answer 1

m = 2kg

R = 0.4305 kJ/kg.K

∆U = 1458.01 kJ

∆H = 2189.13 kJ

$∆H = ∆U -mR∆T\\ 2189.13 = 1458.01 - 2(0.4305)∆T \\ 731.12 =0.861∆T\\ ∆T = 849.15K$