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Solution to Exercise 5.3

9 October, 2015 - 15:37

MATLAB solution is as follows:

    » new_temperature=25:5:150;
» new_specific_heat=interp1(temperature,specific_heat,new_temperature);
» [new_temperature',new_specific_heat']
ans =
 25.0000  0.1180
 30.0000  0.1184
 35.0000  0.1188
 40.0000  0.1192
 45.0000  0.1196
 50.0000  0.1200
 55.0000  0.1206
 60.0000  0.1212
 65.0000  0.1218
 70.0000  0.1224
 75.0000  0.1230  
 80.0000  0.1234
 85.0000  0.1238
 90.0000  0.1242
 95.0000  0.1246
100.0000  0.1250
105.0000  0.1256
110.0000  0.1262
115.0000  0.1268
120.0000  0.1274
125.0000  0.1280
130.0000  0.1286
135.0000  0.1292
140.0000  0.1298
145.0000  0.1304
150.0000  0.1310