Warmtekapasiteit: Verskil tussen weergawes
Content deleted Content added
Lyn 149:
'''Antwoord:'''
<math>C_{p,CO} = 6.6 + 0.0012T \qquad kal/mol \ (T \ in \ Kelvin)</math>
* C<sub>p, CO2</sub> = 10.34 + 0.00274T – 195500/T<sup>2</sup> kal/mol (T in Kelvin)▼
* C<sub>p, H2O(g)</sub>) = 8.22 + 0.00015T + 1.34e-6T<sup>2</sup> kal/mol (T in Kelvin)▼
▲
▲
Die warmtekapasiteit van die mengsel is nou:
<math>C_{p,mengsel} = 0.3C_{p,CO} + 0.6C_{p,CO2} + 0.1C_{p,H2O(g)} = 9.01 + 0.00202T + 1.34 \times 10^{-7}T^2 - \frac{117300}{T^2} \ (T \ in \ Kelvin)</math>
Om die hitteverlies te bepaal, moet hierdie vergelyking nou [[Lys van integrale|geïntegreer]] word:
<math>Q = \int_{T_1}^{T_2} C_p \, dT</math>
Line 163 ⟶ 168:
<math>= \left | 9.01T + \frac{0.00202}{2}T^2 + \frac{1.34 \times 10^{-7}}{3} T^3 - \frac{117300}{-1} T^{-1} \right |_{800}^{400}</math>
<math>= \left [ 9.01(400) + \frac{0.00202}{2}(400)^2 + \frac{1.34 \times 10^{-7}}{3}(400)^3 - \frac{117300}{-1}(400)^{-1} \right ] - \left [ 9.01(800) + \frac{0.00202}{2}(800)^2 + \frac{1.34 \times 10^{-7}}{3}(800)^3 - \frac{117300}{-1}(800)^{-1} \right ]</math>
| |||