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Chapter 5: Problem 24

A 25.0 -mL. sample of benzene at \(19.9^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C},\) and then frozen. How much energy was given off as heat in this process? (The density of benzene is \(0.80 \mathrm{g} / \mathrm{mL},\) its specific heat capacity is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) and its heat of fusion is \(127 \mathrm{J} / \mathrm{g} .\) )

Short Answer

Expert verified
The total energy released is 3040.16 J.

Step by step solution

01

Calculate the Mass of Benzene

First, we need to determine the mass of the benzene sample. Given that the density of benzene is \(0.80 \, \mathrm{g/mL}\) and the volume is \(25.0 \, \mathrm{mL}\), we can use the formula for density: \[\text{Mass} = \text{Density} \times \text{Volume}\]Thus, the mass is:\[0.80 \, \mathrm{g/mL} \times 25.0 \, \mathrm{mL} = 20.0 \, \mathrm{g}\]
02

Calculate the Energy to Cool Benzene

Next, we determine the energy required to cool the benzene from \(19.9^{\circ}\mathrm{C}\) to its melting point of \(5.5^{\circ}\mathrm{C}\). The formula for heat transfer is:\[q = m \cdot c \cdot \Delta T\]where \(q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. Here, \(m = 20.0 \, \mathrm{g}\), \(c = 1.74 \, \mathrm{J/g \cdot K}\), and \(\Delta T = 19.9^{\circ}\mathrm{C} - 5.5^{\circ}\mathrm{C} = 14.4^{\circ}\mathrm{C}\) = \(14.4 \, \mathrm{K}\).Computing, we get:\[q = 20.0 \, \mathrm{g} \times 1.74 \, \mathrm{J/g \cdot K} \times 14.4 \, \mathrm{K} = 500.16 \, \mathrm{J}\]
03

Calculate the Energy to Freeze Benzene

In this step, we calculate the energy absorbed during the phase change from liquid to solid at its melting point. The formula to calculate the energy required for a phase change is:\[q = m \cdot \text{Heat of Fusion}\]Here, \(m = 20.0 \, \mathrm{g}\) and the heat of fusion is \(127 \, \mathrm{J/g}\), so:\[q = 20.0 \, \mathrm{g} \times 127 \, \mathrm{J/g} = 2540 \, \mathrm{J}\]
04

Total Energy Released

The total energy released is the sum of the energy given off while cooling the liquid benzene and the energy released during the freezing process:\[q_{\text{total}} = q_{\text{cool}} + q_{\text{freeze}} = 500.16 \, \mathrm{J} + 2540 \, \mathrm{J}\]Thus, the total energy released is:\[3040.16 \, \mathrm{J}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a property of materials that tells us how much heat energy is needed to raise the temperature of a given amount of a substance by one degree Celsius (or Kelvin). It is an intrinsic property, which means it doesn't change regardless of the size or amount of the substance. In practical terms, a substance with a high specific heat capacity can absorb a lot of heat without changing temperature much. Conversely, substances with low specific heat capacities change temperature significantly even with small amounts of heat.

In our problem, benzene has a specific heat capacity of 1.74 J/g路K. This means that it takes 1.74 Joules of energy to raise 1 gram of benzene by 1 Kelvin. When benzene is cooled from 19.9掳C to 5.5掳C, a 14.4 K decrease, it releases heat instead of absorbing it. We used this property to calculate the energy release during this cooling process. By multiplying the mass of benzene by the specific heat capacity and the temperature change, we find out that 500.16 Joules of energy are released.
Phase Change
A phase change occurs when a substance transitions from one state of matter鈥攕uch as solid, liquid, or gas鈥攖o another. During a phase change, the temperature of the substance remains constant even though it is absorbing or releasing energy. This energy goes into changing the state rather than changing the temperature.

For benzene in our problem, the phase change occurs at the melting point, which is 5.5掳C for benzene. During this process, all the heat energy is used to change the state from liquid to solid. It is important to note that even as heat is removed during freezing, the temperature remains at 5.5掳C until the phase change is complete.

The energy involved in converting a liquid to a solid is calculated using the heat of fusion. This is measured in energy per gram, representing the energy required or released when a gram of the substance undergoes a phase change.
Heat of Fusion
Heat of fusion is the amount of energy needed to change 1 gram of a substance from solid to liquid or vice versa, without a change in temperature. It is a type of latent heat which refers to energy absorbed or released during a phase change without a temperature change.

In the case of benzene, the heat of fusion is given as 127 J/g. This means 127 Joules of energy are released for each gram of benzene as it changes from a liquid to a solid at its melting point. In our exercise, this was the key value used to calculate the energy released during the phase change.

When we calculated the energy released during the freezing of benzene, we multiplied the mass of the benzene sample (20 g) by its heat of fusion (127 J/g). The result was an energy release of 2540 Joules, accounting for the majority of the total energy released as benzene freezes.

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Most popular questions from this chapter

A volume of 1.50 L of argon gas is confined in a cylinder with a movable piston under a constant pressure of \(1.22 \times 10^{5}\) Pa. When \(1.25 \mathrm{kJ}\) of energy in the form of heat is transferred from the surroundings to the gas, the internal energy of the gas increases by 1.11 kJ. What is the final volume of argon gas in the cylinder?

A You have the six pieces of metal listed below, plus a beaker of water containing \(3.00 \times 10^{2} \mathrm{g}\) of water. The water temperature is \(21.00^{\circ} \mathrm{C}.\) $$\begin{array}{|l|c|c|} \hline \text { Metals } & \text { Specific Heat }(J / g K) & \text { Mass }(g) \\\ \hline 1 . A 1 & 0.9002 & 100.0 \\ 2 . A 1 & 0.9002 & 50.0 \\ 3 . A u & 0.1289 & 100.0 \\ 4 . A u & 0.1289 & 50.0 \\ 5 . Z n & 0.3860 & 100.0 \\ 6 . Z n & 0.3860 & 50.0 \\ \hline \end{array}$$ (a) In your first experiment you select one piece of metal and heat it to \(100^{\circ} \mathrm{C},\) and then select a second piece of metal and cool it to \(-10^{\circ} \mathrm{C}\) Both pieces of metal are then placed in the beaker of water and the temperatures equilibrated. You want to select two pieces of metal to use, such that the final temperature of the water is as high as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water? (b) The second experiment is done in the same way as the first. However, your goal now is to cause the temperature to change the least, that is, the final temperature should be as near to \(21.00^{\circ} \mathrm{C}\) as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water?

When \(0.850 \mathrm{g}\) of \(\mathrm{Mg}\) was burned in oxygen in a constant- volume calorimeter, 25.4 kJ of energy as heat was evolved. The calorimeter was in an insulated container with \(750 .\) g of water at an initial temperature of \(18.6^{\circ} \mathrm{C}\). The heat capacity of the bomb in the calorimeter is \(820 . \mathrm{J} / \mathrm{K}.\) (a) Calculate \(\Delta U\) for the oxidation of \(\mathrm{Mg}\) (in kJ/mol Mg). (b) What will be the final temperature of the water and the bomb calorimeter in this experiment?

A piece of chromium metal with a mass of \(24.26 \mathrm{g}\) is heated in boiling water to \(98.3^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing 82.3 g of water at \(23.3^{\circ}\) C. When thermal equilibrium is reached, the final temperature is \(25.6^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of chromium.

As the gas trapped in a cylinder with a movable piston cools, \(1.34 \mathrm{kJ}\) of work is done on the gas by the surroundings. If the gas is at a constant pressure of \(1.33 \times 10^{5} \mathrm{Pa}\), what is the change of volume (in L) of the gas?
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