Question

If a can of soda \(\left(350 \mathrm{~mL} ;\right.\) treat as water) cools from \(20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), how much energy is extracted, and how much is the entropy (in \(\mathrm{J} / \mathrm{K}\) ) in the can reduced using the average temperature and the relation that \(\Delta Q=T \Delta S ?\)

Short answer

Question: Calculate the amount of energy extracted and the reduction in entropy when a 350 mL can of soda cools from 20°C to 0°C. Answer: The energy extracted is 29,260 J, and the reduction in entropy is 103.38 J/K.

Step-by-step solution

Find the energy extracted

To calculate the energy extracted, we'll use the formula: \(Q = m \cdot c \cdot \Delta T\), where \(Q\) is the energy extracted, \(m\) is the mass of the substance, \(c\) is the specific heat capacity and \(\Delta T\) is the change in temperature. For water, the specific heat capacity (c) is approximately \(4.18 \mathrm{J/(g \cdot K)}\). First, we need to find the mass (m) of the soda. The given volume of the soda is 350 mL. Since water has a density of about \(1 \mathrm{g/mL}\), the mass of the soda can be calculated as: \(m = V \cdot \rho= 350 \mathrm{mL} \cdot 1 \mathrm{g/mL} = 350 \mathrm{g}\). Now, we can find the change in temperature (ΔT): \(\Delta T= T_{f} - T_{i} = 0^\circ \mathrm{C} - 20^\circ \mathrm{C} = -20^\circ \mathrm{C}\). From the specific heat formula, we have: \(Q = m \cdot c \cdot \Delta T = (350 \ \text{g}) \cdot (4.18\ \text{ J/(g·K)}) \cdot (-20 \ \text{K}) = -29,\!260\ \text{J}\). The energy extracted is: \(Q = -29,260 \ \text{J}\).

Find the reduction in entropy

To find the reduction in entropy, we'll use the given relationship: \(\Delta Q = T\Delta S\). Since we have already calculated the energy extracted (\(\Delta Q\)), we need to find the average temperature (T) to plug into this formula. The average temperature, in this case, is the average of the initial and final temperatures: \(T = \frac{T_{i}+T_{f}}{2}= \frac{20^\circ \mathrm{C}+0^\circ \mathrm{C}}{2} = 10^\circ \mathrm{C}\). To work with absolute temperatures in Kelvin, we convert the average temperature to K: \(T_K = 10^\circ \mathrm{C} + 273.15 \ \mathrm{K} = 283.15\ \mathrm{K}\). Now, we plug this into the entropy relation: \(-29,260\ \text{J} = (283.15\ \mathrm{K})\Delta S\). Solving for \(\Delta S\), we get: \(\Delta S = \frac{-29,260 \ \text{J}}{283.15\ \mathrm{K}} = -103.38\ \mathrm{J/K}\). The reduction in entropy is: \(\Delta S = -103.38\ \mathrm{J/K}\).

Key concepts

Entropy

Entropy, often denoted as 'S', is a fundamental concept in thermodynamics that measures the level of disorder or randomness in a system. In simple terms, it's a way of quantifying how much energy in a system is not available for doing work because it is randomly distributed among its particles. The second law of thermodynamics states that the entropy in an isolated system always increases over time.
However, entropy can decrease in a localized part of the system if energy is extracted, as seen in our example where cooling the can of soda lowers its entropy. The reduction of entropy is connected with the energy extracted from the system, which questions the soda can’s internal microstates to become more orderly. The relationship between heat transfer (denoted as eq Q) and change in entropy (eq S) is given by the formula eq eq Q = T eq S, with T representing the temperature at which the heat transfer occurs. The higher the temperature, the less impact a certain heat transfer has on the entropy change, while at lower temperatures, even small energy transfers can cause significant shifts in entropy.

Specific Heat Capacity

Specific heat capacity, symbolized as 'c', is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin). Each material has its own specific heat capacity, which is a measure of how much it 'resists' changing its temperature when it absorbs or releases heat. This is why materials with high specific heat capacities, like water, take longer to change temperature compared to materials with lower specific heat capacities.
In the example given, water has a specific heat capacity of about eq 4.18 eq J/(g eq K), which is relatively high. This is why it's often used in heating or cooling systems as it can absorb or release a large amount of heat with little change in its own temperature. In the given exercise, understanding specific heat capacity is crucial for determining the amount of energy needed to lower the temperature of the soda. The calculation of energy transfer involves this constant and highlights its relevance in practical thermodynamic processes.

Temperature Change

Temperature change indicates the difference between the final and initial temperature of a system or object, noted as eq T. It is critical to thermodynamics because it drives the flow of heat; heat moves from a hotter object to a cooler one until thermal equilibrium is reached.

In the context of the example with the soda can, a negative eq T (a drop in temperature) signifies that heat energy leaves the soda, leading to its cooling. This change in temperature is essential to calculating the energy extracted from the soda and subsequently, the consequent changes in the thermodynamic properties of the system, such as entropy. By understanding how to measure and calculate temperature changes, students can apply these principles to various real-world scenarios, from industrial processes to everyday occurrences.

Energy Transfer

Energy transfer is the movement of energy from one place or object to another. It occurs in many forms, such as heat, light, or mechanical work. In thermodynamics, we're typically most interested in heat energy, which is transferred due to a temperature difference and travels from a warmer body to a cooler body until both bodies reach thermal equilibrium.
The energy extracted when the soda cools, as per the problem, is an example of energy transfer. The negative sign in the energy extracted (eq Q = -29,260 eq J) signifies that energy is leaving the substance, in this case, moving out from the soda can into its surroundings, which is usually the refrigerator or the ambient air. As such, this demonstrates a foundational principle in thermodynamics: energy is conserved and can neither be created nor destroyed, only transferred from one form to another.