Problem 46726. [Thermodynamics] Polynomial fitting of heat capacity data

Solution 3199241

Submitted on 13 Oct 2020 by Giovanni Mottola

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Test	Status	Code Input and Output
1	Pass	T = 300:100:1000; % hydrogen @ 300K < T < 1000K CP = [28.85 29.18 29.26 29.32 29.44 29.62 29.88 30.2]; cpfun = cpFitting(T,CP,2); assert(abs(cpfun(350)-29.0074) < 1e-3 && abs(cpfun(550)-29.2505) < 1e-3 && abs(cpfun(940)-29.9943) < 1e-3); P = 0.0000 0.0002 28.8077 cpfun = function_handle with value: @(t)sum(fliplr(P).*t.^(0:N))
2	Pass	T = 500:100:1500; % water/steam @ 500K < T < 1500K CP = [35.22 36.22 37.5 38.74 40 41.27 42.52 43.75 44.94 46.06 47.11]; cpfun = cpFitting(T,CP,3); assert(abs(cpfun(560)-35.842) < 1e-3 && abs(cpfun(1000)-41.26) < 1e-3 && abs(cpfun(1400)-46.0668) < 1e-3); P = -0.0000 0.0000 0.0041 31.2662 cpfun = function_handle with value: @(t)sum(fliplr(P).*t.^(0:N))
3	Pass	T = 100:100:1000; % methane @ 100K < T < 1000K CP = [33.28 33.51 35.76 40.63 46.63 52.74 58.6 64.08 69.14 73.75]; cpfun = cpFitting(T,CP,3); assert(abs(cpfun(290)-35.993) < 1e-3 && abs(cpfun(630)-54.1682) < 1e-3 && abs(cpfun(950)-71.6749) < 1e-3); P = -0.0000 0.0002 -0.0346 35.0203 cpfun = function_handle with value: @(t)sum(fliplr(P).*t.^(0:N))
4	Pass	str = fileread('cpFitting.m'); % sorry, no regexp hacks :-) assert(isempty(regexp(str,'regexp')));

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