Question

Given $C_p = a + bT$
$a = 19.5$, $b = 0.042$, $n = 1$ mole
Temperature changes from $27^\circ C$ to $327^\circ C$.
Find $\Delta H$.

Hint:

If $C_p$ is temperature dependent, always integrate: $\Delta H = \int C_p\, dT$. Remember to convert temperature to Kelvin.

Answer 1

Step 1: Convert temperature to Kelvin.
\[ T_1 = 27^\circ C = 300\,K \] \[ T_2 = 327^\circ C = 600\,K \]
Step 2: Use formula for enthalpy change.
\[ \Delta H = \int_{T_1}^{T_2} C_p \, dT \] \[ = \int_{T_1}^{T_2} (a + bT)\, dT \]
Step 3: Integrate.
\[ \Delta H = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) \]
Step 4: Substitute values.
\[ T_2 - T_1 = 600 - 300 = 300 \] \[ T_2^2 - T_1^2 = 600^2 - 300^2 \] \[ = 360000 - 90000 \] \[ = 270000 \]
Now calculate:
\[ \Delta H = 19.5(300) + \frac{0.042}{2}(270000) \]
\[ = 5850 + (0.021)(270000) \]
\[ = 5850 + 5670 \]
\[ = 11520 \]
Final Answer:
\[ \boxed{\Delta H = 11520} \] (Units depend on given $C_p$, typically J/mol.)