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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Thermodynamics > Calorimetry > Constant Pressure Calorimetry

Constant Pressure Calorimetry

Because calorimetry is used to measure the heat of a reaction, it is a crucial part of thermodynamics. In order to measure the heat of a reaction, the reaction must be isolated so that no heat is lost to the environment. This is achieved by use of a calorimeter, which insulates the reaction to better contain heat. Coffee cups are often used as a quick and easy to make calorimeter for constant pressure. More sophisticated bomb calorimeters are built for use at constant volumes.

Introduction

Using calorimetry, it is possible to calculate the amount of heat gained and lost by each part of the system. The amount of heat lost or gained by any part of the reaction is related to the heat capacity of the substance. Heat capacity is the amount of energy, \(q\), needed to raise the temperature 1˚ C or 1 K. To calculate the amount of heat released or absorbed by a reaction in terms of \(\Delta{T}\), the equation for object heat capacity (below) is used:

\[q = \Delta{T} \times C \tag{1}\]

\(q\)= the amount of heat gained or lost
\(C\) = heat capacity

\(\Delta T \) =  final temperature –  initial temperature

Specific heat capacity takes into account the mass of a substance and the energy needed to raise its temperature by \(\Delta{T}\), as shown in the equation for specific heat capacity below:

\[q =m \times c_{sp} \times \Delta{T} \tag{2}\]

q  = the amount of heat gained or lost
m = mass of the substance
csp = specific heat

\(\Delta T \) = final temperature –  initial temperature

Theoretically, the \(q_{products}\) should be equivalent to \(q_{reactants}\) because the amount of heat gained by one substance should be equal to the amount lost by another. However, the calorimeter also absorbs some of the heat from the reaction. Because of this, the heat capacity of the calorimeter is also necessary for calculations and results in the following equation:

\[q_{reactants}+ q_{products} + q_{calorimeter} = 0 \tag{3}\]

The above equation can also be written as:

qcal = -(qlead + qwater)

Example 1

If 150 g of lead at 100°C were placed in a calorimeter with 50 g of water at 28.8°C and the resulting temperature of the mixture was 22°C, what are the values of qlead, qwater and qcal? (Knowing that the specific heat of water is 4.184 J/g °C and the specific heat of lead is 0.128 J/g °C)

SOLUTION

For lead, we know that:

m(mass) = 150 g,
Ti (initial temperature) = 100°C
Tf = 28.8°C
csp (specific heat of lead) = 0.128 J/g °C

For water:

m= 50 g,
Ti = 22°C
Tf = 28.8°C
csp = 4.184 J/g °C

q =m x csp x \(\Delta T \)

qlead = 0.128 J/g °C x 150g x (28.8°C - 100°C) = -1.37 E3 J
qwater= 4.184 J/g °C x 50g x (28.8°C - 22°C) = 1.42 E3 J

qcal = -(qlead + qwater)

qcal = -(qlead + qwater) = -(1.42E3 +-1.37 E3) = -50.0 J

In a Coffee-Cup Calorimeter, reactants are mixed in a solution, usually water, and measure the temperature change. Since styrofoam acts as a very effective insulator, little heat transfer takes place between the cup and the surrounding air. Everything within the cup, including the cup itself, is considered an isolated system. The heat of reaction is clearly defined as "the quantity of heat that would be exchanged with the surroundings in restoring the calorimeter to its initial temperature." But the calorimeter cannot be restored back to its original conditions, which is why we take the negative quantity of heat producing the temperature change in the calorimeter.

\(q_{reaction} + q_{calorimeter} = 0 \)

Which means,

\(q_{reaction} = - q_{calorimeter} \)

Note that, 

\(q = mass \; \times \; specific \; heat \; \times \; temperature \; change\)

Temperature can be in Kelvin or Celsius, and the change in temperature, \(\Delta T \), is calculated by taking \(T_{final} - T_{initial}\).

 

Example 2: I see you

When 20g of Al (s) at 98 degrees Celcius is placed in 50g of H2O, the final temperature is 26.5 degrees Celcius. What was the initial temperature of H2O? Given: CspAl(s) = 0.903 J/g oC and CspH20(l) = 4.18 J/g oC

SOLUTION

\(q_{H_2O} = - q_{Al}\)

\((m_{H_2O}) \; (C_{spH_2O}) \; (\Delta{T}) = (m_{Al}) \; (C_{spAl}) \; (\Delta{T})\)

\((50 \;g)(4.18 \; J/g^oC)(26.5^oC - T_{iH_2O}) = -(20 \; g)(0.903 \; J/g^oC)(26.5 - 98^oC)\) 

Using algebra, solve for \(T_{iH_2O}\):

\[T_{iH_2O} = 20^oC\]

 

Example 3

A 28.2 gram sample of Copper (Csp Cu= 0.685 J/g K) is placed in a coffee cup calorimeter containing 100 grams of water that just stopped boiling. After some time the temperature of the water becomes constant at 92.3 oC. Assuming the atmospheric pressure is 1 atmosphere, calculate the initial temperature of the Copper block. Assume no heat is lost to the surroundings. Csp H2O(l) = 4.18 J/g oC)

SOLUTION

\(q_{H_2O} = - q_{Cu}\)

\((m_{H_2O}) \; (C_{spH_2O}) \; (\Delta{T}) = (m_{Cu}) \; (C_{spCu}) \; (\Delta{T})\)

\((100 \;g)(4.18 \; J/g^oC)(-7.7^oC) = -(28.2 \; g)(0.685 \; J/g^oC)(92.3^oC - T_{f \; Cu})\) 

Solve for \(T_{f \; Cu}\):

\(-3218.6 = -19.317 (92.3 - T_{f \; Cu})\)

\(1435.64 = 19.32 \; T_{f \; Cu}\)

\(T_{f \; Cu} = 74.32 = 74.3^oC\)

Note: Notice the difference in temperature units given by the two Csp values. The Csp Cu = 0.685 J/g K is given in Kelvins and the CspH2O = 4.18 J/g oC is given in Celcius. There is no need for conversions because when Csp values are given, the change in temperature will remain constant. In other words, no matter what the units are, the change will still occur and be exactly the same as would naturally.

References

  1. Harwood , William S., F. Geoffrey Herring, Ralph H. Petrucci, and Jeffry D. Madura. General Chemistry: Principles and Modern Applications. 9th ed.. Upper Saddle River, NJ: Pearson Education, inc., 2007.

Contributors

  •  Michelle Dube, Allison Billings (UCD), Rachel Morris (UCD), Ryan Starr (UCD)

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Last Modified
15:00, 25 Jun 2014

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