Question

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{kJ}\) and \(w=-840 \mathrm{J} .\) (b) A system releases 66.1 \(\mathrm{kJ}\) of heat to its surroundings while the surroundings do 44.0 \(\mathrm{kJ}\) of work on the system.

Short answer

For case (a), the change in internal energy is \(\Delta E = -0.077 \mathrm{kJ}\), and the process is endothermic. For case (b), the change in internal energy is \(\Delta E = -22.1 \mathrm{kJ}\), and the process is exothermic.

Step-by-step solution

Case (a) - Calculate \(\Delta E\)

First, we need to convert the units of work to be consistent with the units of heat transfer. We'll convert work done from Joules (J) to kilojoules (kJ) using the conversion factor, \(1 \mathrm{kJ} = 1000 \mathrm{J}\): $$w = -840 \mathrm{J} \times \frac{1 \mathrm{kJ}}{1000 \mathrm{J}} = -0.840 \mathrm{kJ}$$ Now, we can calculate the change in internal energy using the formula \(\Delta E = q + w\): $$\Delta E = 0.763 \mathrm{kJ} + (-0.840 \mathrm{kJ}) = -0.077 \mathrm{kJ}$$

Case (a) - Determine endothermic or exothermic

Since the heat transfer \(q = 0.763 \mathrm{kJ}\) is a positive value, it means that the process is endothermic, as the system absorbed heat.

Case (b) - Calculate \(q\) and \(w\)

In this case, we are given that the system releases \(66.1 \mathrm{kJ}\) of heat and the surroundings do \(44.0 \mathrm{kJ}\) of work on the system. Thus, $$q = -66.1 \mathrm{kJ}$$ $$w = 44.0 \mathrm{kJ}$$

Case (b) - Calculate \(\Delta E\)

We can calculate the change in internal energy using the formula \(\Delta E = q + w\): $$\Delta E = -66.1 \mathrm{kJ} + 44.0 \mathrm{kJ} = -22.1 \mathrm{kJ}$$

Case (b) - Determine endothermic or exothermic

Since the heat transfer \(q = -66.1 \mathrm{kJ}\) is a negative value, it means that the process is exothermic, as the system released heat to the surroundings. In conclusion: - For case (a), the change in internal energy is \(\Delta E = -0.077 \mathrm{kJ}\), and the process is endothermic. - For case (b), the change in internal energy is \(\Delta E = -22.1 \mathrm{kJ}\), and the process is exothermic.

Key concepts

Understanding Endothermic and Exothermic Processes

In thermodynamics, the energy changes occurring within a system are of paramount importance. Among these, endothermic and exothermic processes are two fundamental concepts that are often encountered.

An endothermic process occurs when a system absorbs heat from its surroundings, resulting in a positive heat transfer value, denoted as 'q'. This process usually leads to a decrease in the temperature of the surroundings as energy is taken in. In contrast, an exothermic process releases heat to its surroundings, which is indicated by a negative 'q' value, resulting in an increase in the surrounding temperature.

For example, when a chemical reaction requires heat from the environment to proceed, such as ice melting or photosynthesis, it's an endothermic reaction. Conversely, combustion or the freezing of water are exothermic since they release heat. The exercise example for case (a), with \(q = 0.763 \mathrm{kJ}\), implies that the system absorbed heat, making the process endothermic. Conversely, case (b) is an example of an exothermic process where the system loses heat, evident from \(q = -66.1 \mathrm{kJ}\).

Heat Transfer in Thermodynamics

Heat transfer is a key concept in thermodynamics as it relates to the movement of thermal energy between physical systems. Heat can move via conduction, convection, and radiation, and does not require any contact between the heat source and the object being heated in all cases. The direction of heat transfer is always from the hotter object to the cooler one until thermal equilibrium is reached.

In the context of the provided exercise, understanding heat transfer is crucial. Case (a) shows 0.763 kJ of heat entering the system, whereas in case (b), the system surrenders 66.1 kJ to the surroundings, demonstrating a transfer of heat away from the system. When addressing homework problems like this, it's important to note that heat transfer values can directly indicate whether a process is endothermic (heat entering the system) or exothermic (heat leaving the system). The key is to remember positive values for 'q' indicate heat coming in, while negative values mean heat going out.

Work Done in Thermodynamics

In thermodynamics, work is defined as the energy transferred from one system to another that is not heat. Work can be done on or by the system. For instance, when a gas expands in a cylinder, it pushes against the piston, doing work on the surroundings. Conversely, work is done on the gas when the piston compresses it.

The sign conventions for work done can be confusing, but a simple rule applies: if the system does work on the surroundings, work (w) is negative; if work is done on the system, w is positive. This convention assumes that we are considering the system's perspective. For example, the exercise indicates that in case (a), the work done by the system, represented by \(w = -0.840 \mathrm{kJ}\), shows a negative value because the system has performed work on its surroundings. In case (b), work is done on the system \(w = 44.0 \mathrm{kJ}\), which is positive since the surroundings are doing work on the system. This adjustment to internal energy, \(\Delta E\), highlights the dual aspects of energy change: heat and work, both contributing to the overall internal energy change.