Question

Some of the butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) in a \(200.0 \mathrm{L}\) cylinder at \(26.0^{\circ} \mathrm{C}\) is withdrawn and burned at a constant pressure in an excess of air. As a result, the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 L of water in a heater from 26.0 to 62.2 ^ C. Assume that the combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) exclusively, and determine the efficiency of the water heater. (That is, what percent of the heat of combustion was absorbed by the water?)

Short answer

To obtain the short answer, follow the above steps to calculate the efficiency of the water heater as a percentage.

Step-by-step solution

Calculate the amount of butane consumed

Using the ideal gas law \(PV = nRT\), we can calculate the number of moles of butane consumed. Initially, the number of moles is given by \(\frac{{2.35 \times 200.0}}{{0.0821 \times (26+273)}}\). After burning, the number of moles is \(\frac{{1.10 \times 200.0}}{{0.0821 \times (26+273)}}\). The difference between these two values gives the number of moles of butane consumed.

Heat liberated by butane combustion

The heat of combustion of butane is −2878 kJ/mol (This value is usually given in the problem or can be looked up in a chemistry reference book). The total heat liberated is the heat of combustion times the number of moles of butane consumed.

Heat absorbed by water

The heat absorbed by the water can be calculated by using the formula \(q = mc\Delta T\), where m is the mass of water, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the temperature change. Here, the volume of water is given, so we can use the fact that 1 L of water has a mass of 1 kg, so the mass of water being heated is 132.5 kg. The specific heat capacity of water is 4.184 J/g°C, and the temperature change is from 26.0°C to 62.2°C.

Calculate the efficiency

Efficiency is the ratio of the useful output to the total input expressed in percentage. Here, the useful output is the heat absorbed by the water and the total input is the heat liberated by the butane combustion. Hence, the efficiency is \(\frac{{\text{{Heat absorbed by water}}}}{{\text{{Total heat liberated}}}} \times 100\%\)

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that describes the behavior of ideal gases. It is represented by the formula \( PV = nRT \).
Here, \( P \) denotes the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (approximately 0.0821 L·atm/mol·K), and \( T \) represents the temperature in Kelvin.

• To convert the Celsius temperature to Kelvin, you simply add 273.15 to the Celsius temperature.
• This equation allows us to calculate unknown quantities such as the number of moles if other values are known.
• In our exercise, we used this law to calculate how many moles of butane were consumed by the change in pressure after combustion.

Understanding the Ideal Gas Law helps us determine the behavior and properties of gases under varying conditions of pressure, volume, and temperature.

Heat of Combustion

The heat of combustion is the energy released when a substance completely burns in oxygen. For butane ( C_4H_{10} ), the heat of combustion is given as -2878 kJ/mol.
This value indicates the amount of energy produced when one mole of butane is burned.

• Negative sign: The negative value shows that the combustion process is exothermic, meaning it releases energy.
• This energy can be transferred to surrounding systems, like heating water in our example.
• The total heat liberated during the combustion is calculated by multiplying the heat of combustion by the number of moles of butane consumed.

Knowing the heat of combustion is essential for calculating energy efficiency, as it tells us how much energy is available to be harnessed in useful ways.

Specific Heat Capacity

Specific heat capacity refers to the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is 4.184 J/g°C.

• It indicates the energy needed to change the temperature of water, facilitating the calculation of energy absorption during heating.
• To find the heat absorbed by the water, the formula \( q = mc\Delta T \) is used. Here, \( m \) represents mass, \( c \) is the specific heat capacity, and \( \Delta T \) indicates the change in temperature.
• In our case, the mass of water is 132.5 kg, which translates to 132,500 grams, given that 1 liter of water equals 1 kg.

Understanding specific heat capacity allows us to predict how substances will respond to energy transfer in terms of heat.

Energy Transfer

Energy transfer during combustion is a pivotal concept in this problem. The energy released from burning butane is transferred as heat to the water in a heater.
The efficiency of this energy transfer involves comparing the useful energy harnessed (absorbed by the water) to the total energy released by combustion.

• Calculate the efficiency using the formula \( \frac{\text{Heat absorbed by water}}{\text{Total heat liberated}} \times 100\% \).
• It measures how much of the initial energy ends up being used effectively, in this case, heating the water.
• Inefficiencies can result from heat loss to the environment and other system limitations.

By analyzing energy transfer, we can gain insights into the practical use of energy systems and the effectiveness of processes involving combustion.