What is the final temperature when heat in the amount of 400 × 10^6 J is added to 11 × 10³ mol of ammonia initially at 530 K in a steady-flow process at 1 bar?

Question

What is the final temperature when heat in the amount of [latex] 400 imes 10^{6} J[/latex] is added to 11 × 10³ mol of ammonia initially at 530 K in a steady-flow process at 1 bar?

Accepted Answer

If ΔH is the enthalpy change for 1 mol, Q = n ΔH, and

[latex]\Delta H = \frac{\varrho }{n} = \frac{400 imes 10^{6} }{11,000} =  36,360 J· mol^{−1}[/latex]

Then for any value of T, with parameters from Table C.1 and [latex]R = 8.314 J· mol^{−1}·K^{−1}[/latex]:

[latex] \frac{\left\langle C_{P} ight angle_{H} }{R}= MCPH (530, T; 3.578, 3.020  imes   10^{-3} , 0.0, −0.186  imes 10^{5} ) [/latex]

This equation and Eq. (4.11) together can be solved for T, yielding T = 1234 K.

[latex]T = \frac{\Delta H}{\left\langle C_{P} ight angle _{H} } +T_{0} [/latex] (4.11)

A trial procedure is an alternative approach to solution of this problem. One sets up an equation for Q by combining Eqs. (4.3) and (4.8),

[latex]\varrho = \Delta H = \int_{T_{1} }^{T_{2}}{C_{P} dt }[/latex] (4.3)

[latex]\int_{T_{0}}^{T }{\frac{C_{P} }{R} dT} =A (T-T_{0}) + \frac{B}{2}(T^{2} - T^{2}_{0} ) + \frac{C}{3} (T^{3} - T^{3}_{0})+ D \left(\frac{T-T_{0}}{TT_{0}} ight)[/latex] (4.8)

with T as an unknown on the right. With Q known, one merely substitutes a rational succession of values for T until the known value of Q is reproduced. Microsoft Excel’s Goal Seek functionis an example of an automated version of this procedure.

Question 4.3: What is the final temperature when heat in the amount of 400...

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