Question

A system undergoes a thermodynamic cycle while receiving energy by heat transfer from a tank of liquid water initially at \(90^{\circ} \mathrm{C}\) and rejecting energy by heat transfer at \(15^{\circ} \mathrm{C}\) to the surroundings. If the final water temperature is \(15^{\circ} \mathrm{C}\), determine the minimum theoretical volume of water in the tank, \(\mathrm{m}^{3}\), for the cycle to produce net work equal to \(1.6 \times 10^{5} \mathrm{~kJ}\).

Short answer

The minimum theoretical volume of water is approximately 2.471 m³.

Step-by-step solution

- Identify key information

Extract the given values from the problem. \(T_{h} = 90^{\bullet}C = 363.15 \ K\), \(T_{c} = 15^{\bullet}C = 288.15 \ K\), final temperature of water is \(15^{\bullet}C\), net work \(W = 1.6 \times 10^{5} \ kJ\).

- Calculate efficiency using Carnot cycle

The efficiency of a Carnot cycle is given by \(\eta = 1 - \frac{T_{c}}{T_{h}}\) Plug in the temperatures \(\eta = 1 - \frac{288.15}{363.15}\) \(\eta \approx 0.2067\).

- Calculate the required heat input

The net work output is related to the heat input by the efficiency: \(W = \eta Q_{h}\) Rearrange to solve for \(Q_{h}\): \(Q_{h} = \frac{W}{\eta}\) Plug in the values: \(Q_{h} = \frac{1.6 \times 10^{5} \ kJ}{0.2067}\) \(Q_{h} \approx 774,452.8 \ kJ\).

- Relate heat transfer to mass of water

The specific heat of water \(c = 4.18 \ \mathrm{kJ/kg⋅K}\). The energy required to change the temperature of water is \(Q = m c \Delta T\), where \(\Delta T = T_{h} - T_{c}\) \(\Delta T = 363.15 \ K - 288.15 \ K = 75 \ K\). Rearrange to solve for mass \(m\): \(m = \frac{Q_{h}}{c \Delta T}\) Plug in the values: \(m \approx \frac{774,452.8 \ kJ}{4.18 \ kJ/kg⋅K \times 75 \ K}\) \(m \approx 2471 \ kg\).

- Convert mass to volume

The density of water is \(1000 \ \mathrm{kg/m^3}\). The volume \(V\) is given by \(V = \frac{m}{\text{density}}\) \(V = \frac{2471 \ kg}{1000 \ kg/m^3}\) \(V \approx 2.471 \ m^3\).

Key concepts

Carnot Cycle Efficiency

The Carnot cycle is an idealized thermodynamic cycle that defines the maximum possible efficiency any heat engine can achieve. This efficiency is dependent on the temperatures of the heat source and heat sink.
The formula for Carnot efficiency is \(\text{Efficiency}_{\text{Carnot}} = 1 - \frac{T_{c}}{T_{h}}\), where \(T_{h}\) is the absolute temperature of the heat source (hot reservoir) and \(T_{c}\) is the absolute temperature of the heat sink (cold reservoir).
This cycle sets a theoretical upper limit on efficiency, meaning no real engine operating between these two temperatures can exceed this efficiency. From the given problem, our temperatures were: \(T_h = 363.15 \ K\) and \(T_c = 288.15 \ K\). After plugging in these values, the efficiency was calculated to be approximately 0.2067, or 20.67%. This tells us that only 20.67% of the heat absorbed from the hot reservoir is converted into useful work.

Heat Transfer

Heat transfer is the process by which thermal energy moves from a hotter substance to a cooler one. In thermodynamic systems, heat transfer can occur in various ways, such as conduction, convection, and radiation.
For this particular problem, heat transfer is calculated using the efficiency obtained from the Carnot cycle. The relationship between the net work produced by the cycle \(W\) and the heat input \(Q_h\) is given by \(W = \text{Efficiency} \times Q_h\).
By rearranging this formula to solve for \(Q_h\), we find that \(Q_h = \frac{W}{\text{Efficiency}}\). Given the net work output \(W = 1.6 \times 10^5 \ kJ\) and the Carnot efficiency of 0.2067, the heat input required was calculated to be approximately 774,452.8 kJ. This represents the energy needed from the heat source.

Specific Heat Capacity

Specific heat capacity (c) is a property of a substance that indicates the amount of heat required to change the temperature of one kilogram of that substance by one degree Kelvin or Celsius. Water, for example, has a specific heat capacity of 4.18 kJ/kg·K.
The formula to calculate the heat required to change the temperature is \(Q = mc\text{ } \Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \( \Delta T \) is the temperature change.
In our problem, the temperature change \( \Delta T \) was from 90°C to 15°C, or 75 K in absolute terms. Given a required heat input \(Q_h\) of 774,452.8 kJ and the known specific heat capacity of water, we used the formula to solve for the mass \(m\). Plugging the values in, we determined the mass to be approximately 2471 kg. This tells us how much water is necessary to absorb this much thermal energy for the given temperature change.

Temperature Change

Temperature change (\(\Delta T\)) is a critical factor in thermodynamics as it drives the heat transfer process. It is defined as the difference between the final and initial temperatures of a substance.
In this particular problem, the water's temperature changes from 90°C (363.15 K) to 15°C (288.15 K), making the temperature difference \(\Delta T\) equal to 75 K.
This temperature change causes the water to absorb a certain amount of heat, which is essential for the thermodynamic cycle to produce the desired net work. Since the specific heat capacity of water is known, the amount of heat energy absorbed by a given mass of water can be calculated. The energy balance equation \(Q = mc\Delta T\) is then used to determine the total heat transfer required for this temperature change, which in turn reveals the energy dynamics in the thermodynamic cycle.