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Solution to Exercise 5.2

9 October, 2015 - 15:37

MATLAB solution is as follows:

    » temperature=[25;50;75;100;125;150]
 
temperature =
 
    25
    50
    75
    100
    125
    150
 
» specific_heat=[.118;.120;.123;.125;.128;.131]
 
specific_heat =
 
    0.1180
    0.1200
    0.1230
    0.1250
    0.1280
    0.1310
» specific_heatAt30=interp1(temperature,specific_heat,30)
specific_heatAt30 =
    0.1184
» specific_heatAt70=interp1(temperature,specific_heat,70)
specific_heatAt70 =
    0.1224
» specific_heatAt145=interp1(temperature,specific_heat,145)
specific_heatAt145 =
    0.1304