Question

A \(1200-W\) electric resistance heating element whose diameter is \(0.5 \mathrm{cm}\) is immersed in \(40 \mathrm{kg}\) of water initially at \(20^{\circ} \mathrm{C} .\) Assuming the water container is well-insulated, determine how long it will take for this heater to raise the water temperature to \(50^{\circ} \mathrm{C}\). Also, determine the entropy generated during this process, in \(\mathrm{kJ} / \mathrm{K}\).

Short answer

Answer: The time it takes for the heater to raise the water temperature to 50°C is 4180 seconds, and the entropy generated during the process is 13.81 kJ/K.

Step-by-step solution

Calculate the energy needed to raise the water temperature

First, we need to determine the amount of energy (heat) required to raise the temperature of the water. We can do this using the formula for heat transfer: \(Q = mc\Delta T\) where \(Q\) is the heat energy, \(m\) is the mass of water, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the change in temperature. The specific heat capacity of water is 4.18 kJ/(kg·K). The given initial and final temperatures are 20°C and 50°C, respectively: \(\Delta T = T_{final} - T_{initial} = 50°C - 20°C = 30°C\) Now, we can calculate the heat energy: \(Q = (40\,\text{kg}) \times (4.18\,\text{kJ}/(\text{kg}\cdot\text{K})) \times (30\,\text{K}) = 5016\,\text{kJ}\)

Calculate the time required for the heater to transfer this energy

We are given the power rating of the heating element: \(1200\,\text{W}\), which is equivalent to \(1.2\,\text{kW}\). The power tells us how much energy is provided per unit time: \(P=\frac{Q}{t}\), rearranging the relation for time we get \(t = \frac{Q}{P} = \frac{5016\,\text{kJ}}{1.2\,\text{kW}} = 4180\,\text{s}\)

Calculate the change in entropy of the water

To find the change in entropy of the water, we use the specific heat capacity at constant pressure, and the equation: \(\Delta S = mc\ln\left(\frac{T_2}{T_1}\right)\) First, we convert the given temperatures to Kelvin: \(T_1 = 20°C + 273 = 293\,\text{K}\) \(T_2 = 50°C + 273 = 323\,\text{K}\) Now, we calculate the change in entropy: \(\Delta S = (40\,\text{kg}) \times (4.18\,\text{kJ}/(\text{kg}\cdot\text{K})) \times \ln\left(\frac{323\,\text{K}}{293\,\text{K}}\right) = 126.15\,\text{kJ/K}\)

Calculate the entropy generated during the process

We have found the change in entropy of the water, \(\Delta S\). The entropy generated during the process can be found using the formula: \(\Delta S_{gen} = \Delta S - \frac{Q}{T}\) We use the average temperature of the water as the temperature at which the heat is transferred: \(T_{ave} = \frac{T_1 + T_2}{2} = \frac{293\,\text{K} + 323\,\text{K}}{2} = 308\,\text{K}\) Now, we calculate the entropy generated: \(\Delta S_{gen} = 126.15\,\text{kJ/K} - \frac{5016\,\text{kJ}}{308\,\text{K}} = 13.81\,\text{kJ/K}\) The time it takes for the heater to raise the water temperature to 50°C is 4180 seconds, and the entropy generated during this process is 13.81 kJ/K.

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics, playing a vital role in understanding how energy in the form of heat moves from one place to another. It occurs through three main mechanisms: conduction, convection, and radiation. In the context of our exercise, heat is transferred from the electric resistance heating element to the water via conduction.

Since the water container is assumed to be well-insulated, we can neglect heat loss to the surroundings. This assumption allows us to calculate the amount of energy needed to raise the temperature of the water simply by considering the heat provided by the heater. The power rating of the heating element, given in watts, tells us the rate at which the element can transfer energy to the water. Applying the formula for heat transfer, we determine how much energy is required to achieve the desired increase in temperature.

Specific Heat Capacity

The specific heat capacity, often symbolized as 'c', is indicative of the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius (or Kelvin). In our exercise, the specific heat capacity of water, which is 4.18 kJ/(kg·K), demonstrates water's relatively high capacity to absorb heat without experiencing a significant change in temperature.

Understanding specific heat capacity is crucial because it helps predict the amount of energy input needed to achieve a certain temperature change in a material. With water's specific heat capacity known, we can calculate the total heat energy (Q) needed to raise 40 kg of water from 20°C to 50°C by multiplying it with the mass of the water and the temperature change. This concept is especially relevant in designing systems involving heating or cooling where the energy requirements depend on the properties of the substances being heated or cooled.

Entropy Change Calculation

The calculation of entropy change, denoted as \(\Delta S\), gives us insight into the disorder, or randomness, of a system's molecules as it undergoes a process. In thermodynamics, the principle of entropy is tied to the second law, which states that entropy of an isolated system never decreases over time.

In the given exercise, we use the specific heat capacity and the natural logarithm of the ratio of the final and initial temperatures to calculate the change in entropy during the heating process. Recognizing that the entropy change is a measure of energy dispersal at a particular temperature, we consider the effect of an absolute temperature scale by converting Celsius to Kelvin. Once we have calculated the total change in entropy of the water, we find the entropy generated due to the process. The entropy generated is not only a signature of energy transformation but also represents the irreversibility associated with the heating process, emphasizing why it's an essential component in calculating the total change in a system's entropy.