## Integration of the heat capacity

Posted September 05, 2018 at 01:48 PM | categories: uncategorized | tags:

## Table of Contents

From thermodynamics, the heat capacity is defined as \(C_p = \left(\frac{dH}{dT}\right)_P\). That means we can calculate the heat required to change the temperature of some material from the following integral:

\(H_2 - H_1 = Q = \int_{T_1}^{T_2} C_p(T) dT\)

In the range of 298-1200K, the heat capacity of CO_{2} is given by a Shomate polynomial:

\(C_p(t) = A + B t + C t^2 + D t^3 + E/t^2\) with units of J/mol/K.

where \(t = T / 1000\), and \(T\) is the temperature in K. The constants in the equation are

value | |
---|---|

A | 24.99735 |

B | 55.18696 |

C | -33.69137 |

D | 7.948387 |

E | -0.136638 |

F | -403.6075 |

G | 228.2431 |

H | -393.5224 |

## 1 Integrate the heat capacity

Use this information to compute the energy (Q in kJ/mol) required to raise the temperature of CO_{2} from 300K to 600K. You should use `scipy.integrate.quad`

to perform the integration.

### 1.1 solution solution

A = 24.99735 B = 55.18696 C = -33.69137 D = 7.948387 E = -0.136638 F = -403.6075 G = 228.2431 H = -393.5224 def Cp(T): t = T / 1000 return A + B*t + C*t**2 + D*t**3 + E / t**2 from scipy.integrate import quad dH, _ = quad(Cp, 300, 600) print(f'The change in enthalpy is {dH / 1000:1.3f} kJ/mol')

The change in enthalpy is 12.841 kJ/mol

## 2 Verify via Δ H

The change in enthalpy (in kJ / mol) from standard state is

\(dH − dH_{298.15}= A t + B t^2/2 + C t^3/3 + D t^4/4 − E/t + F − H\)

again, \(t = T / 1000\).

Use this equation to compute the change in enthalpy when you increase the temperature from 300 K to 600 K.

### 2.1 solution solution

def dH(T): t = T / 1000 return A * t + B*t**2 / 2 + C * t**3 / 3 + D * t**4 / 4 - E/t + F - H print(f'The change in enthalpy is {dH(600) - dH(300):1.3f} kJ/mol')

The change in enthalpy is 12.841 kJ/mol

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