Question

Water is boiled at sea level in a coffee maker equipped with a 20 -cm-long \(0.4\)-cm-diameter immersion-type electric heating element made of mechanically polished stainless steel. The coffee maker initially contains \(1 \mathrm{~L}\) of water at \(14^{\circ} \mathrm{C}\). Once boiling starts, it is observed that half of the water in the coffee maker evaporates in \(25 \mathrm{~min}\). Determine the power rating of the electric heating element immersed in water and the surface temperature of the heating element. Also determine how long it will take for this heater to raise the temperature of \(1 \mathrm{~L}\) of cold water from \(14^{\circ} \mathrm{C}\) to the boiling temperature.

Short answer

The power rating of the electric heating element is 7.523 kW. 2. What is the surface temperature of the heating element? The surface temperature of the heating element is 100.096°C. 3. How long does it take for the heater to raise the temperature of 1 L of water from 14°C to the boiling temperature? It takes 48.74 minutes to raise the temperature of 1 L of water from 14°C to the boiling temperature.

Step-by-step solution

Calculate the amount of evaporated water and mass flow rate of evaporation.

We are given that half of the initial water in the coffee maker evaporates in 25 minutes. So, the mass of evaporated water is: $$ m_e = 0.5 \times 1 \mathrm{kg} = 0.5 \mathrm{kg} $$ The mass flow rate of evaporation is: $$ \dot{m}_e = \frac{m_e}{t} = \frac{0.5 \mathrm{kg}}{25 \times 60 \mathrm{s}} = \frac{1}{300} \frac{\mathrm{kg}}{\mathrm{s}} $$

Calculate the power required for water evaporation.

The power required for the evaporation process is given by: $$ \dot{Q}_e = \dot{m}_e \times h_{fg} $$ where \(h_{fg}\) is the latent heat of vaporization for water at the boiling point. At sea level, \(h_{fg} = 2256.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\). Therefore, the power required for water evaporation is: $$ \dot{Q}_e = \frac{1}{300} \frac{\mathrm{kg}}{\mathrm{s}} \times 2256.9 \frac{\mathrm{kJ}}{\mathrm{kg}} = 7.523 \mathrm{kW} $$

Calculate the surface area of the heating element.

We know the length and diameter of the heating element. Hence, we can calculate the surface area, \(A_s\), of the heating element: $$ A_s = \pi D L = \pi \times 0.004 \mathrm{m} \times 0.2 \mathrm{m} = 0.002513 \mathrm{m^2} $$

Calculate the convective heat transfer coefficient, \(h\).

From the given condition, the polished stainless steel heating element has an emissivity value, \(\varepsilon = 0.05\). Also, the boiling water temperature is \(100^{\circ} \mathrm{C}\) or \(373 \mathrm{K}\). Therefore, the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m^2 K^4}}\). We can calculate the radiative heat flux, \(q_r\), using the following formula: $$ q_r = \varepsilon \sigma (T_s^4 - T_{w}^4) $$ Now, the convective heat transfer coefficient, \(h\), can be determined by knowing that the total heat flux, \(q_t\), is equal to the sum of the radiative and convective heat flux: $$ q_t = h (T_s - T_w) + q_r $$ The total heat flux can be found using the power required for evaporation and surface area: $$ q_t = \frac{\dot{Q}_e}{A_s} = \frac{7.523 \mathrm{kW}}{0.002513 \mathrm{m^2}} = 2.99 \times 10^6 \frac{\mathrm{W}}{\mathrm{m^2}} $$ By using the formula for \(q_t\), the convective heat transfer coefficient \(h\) can be calculated, which comes around \(h = 1.197 \times 10^6 \frac{\mathrm{W}}{\mathrm{m^2 K}}\).

Find the surface temperature of the heating element and the power rating.

Now, we can use the formula for the convective heat flux, \(q_c = h (T_s - T_w)\), to find the surface temperature of the heating element, \(T_s\). From the formula for \(q_t\): $$ T_s = \frac{q_t - q_r}{h} + T_w $$ The surface temperature of the heating element comes around \(373.246 \mathrm{K}\) or \(100.096^{\circ} \mathrm{C}\). Finally, the power rating of the electric heating element is equal to the power required for evaporation, which is: $$ P = \dot{Q}_e = 7.523 \mathrm{kW} $$

Determine how long it takes for the heater to raise the temperature.

The time required to heat \(1 \mathrm{L}\) of water from \(14^{\circ} \mathrm{C}\) to boiling point can be calculated using the energy conservation equation: $$ P \times t = m \times C_p \times (T_{boil} - T_{initial}) $$ where \(P = 7.523 \mathrm{kW}\) is the heater power, \(m = 1 \mathrm{kg}\) is the mass of water, \(C_p = 4.186 \frac{\mathrm{kJ}}{\mathrm{kgK}}\) is the specific heat of water, \(T_{boil} = 100^{\circ} \mathrm{C}\), and \(T_{initial} = 14^{\circ} \mathrm{C}\). Solving for \(t\) yields: $$ t = \frac{m \times C_p \times (T_{boil} - T_{initial})}{P} = \frac{1 \mathrm{kg} \times 4.186 \frac{\mathrm{kJ}}{\mathrm{kgK}} \times (100 - 14)}{7.523 \mathrm{kW}} = 48.74 \mathrm{min} $$ In summary, the power rating of the electric heating element immersed in water is \(7.523 \mathrm{kW}\), the surface temperature of the heating element is \(100.096^{\circ} \mathrm{C}\), and it takes \(48.74 \mathrm{min}\) to raise the temperature of \(1 \mathrm{L}\) of water from \(14^{\circ} \mathrm{C}\) to the boiling temperature.

Key concepts

Mass Transfer

Mass transfer refers to the movement of a substance from one location to another. In the context of the exercise, it involves the evaporation of water in the coffee maker.
During this process, half of the water, which is initially present, changes from a liquid state to a vapor state and escapes into the atmosphere.
This is a phase change that requires the breaking of molecular bonds, which in turn requires energy.The rate at which this mass transfer of water occurs is critical for designing heating systems.
The mass flow rate is a measure of how much mass is being transferred per unit of time. It can be expressed using the formula:

• Mass flow rate ( \( \dot{m}_e \) ) = mass of evaporated water ( \( m_e \) ) / time ( \( t \) )

In our exercise, half of the water evaporates in 25 minutes, giving a mass flow rate that helps in determining the power required to sustain the boiling process.

Boiling Process

The boiling process is an efficient way to transfer heat energy into a fluid to cause a phase change from liquid to vapor.
At sea level, water boils at 100°C, where sufficient energy is supplied continuously to keep the fluid at its boiling point.
The energy needed for this process primarily comes from an electric heating element in the coffee maker.The latent heat of vaporization, \( h_{fg} \), is a vital aspect of boiling, representing the amount of energy required to transform a kilogram of liquid water into vapor at the boiling point.
For water, this is a significant amount of energy, requiring 2256.9 kJ/kg at sea level.
This latent heat must be provided to maintain boiling without increasing the temperature of the water further.Understanding the boiling process helps in optimizing the design of heating elements in appliances where boiling is critical.
Efficient designs ensure minimum energy wastage and quicker operations.

Electric Heating Element

An electric heating element is a device that converts electrical energy into heat energy.
This piece of equipment is crucial for applications like boiling water in a coffee maker.
The heating element's power rating is a measure of its capability to transfer heat energy over a period.In our exercise, the heating element ensures the continuous supply of energy to the water.
It maintains the boiling point once the water reaches it and facilitates the evaporation process efficiently.
The surface temperature of the heating element slightly exceeds the boiling point of water to ensure efficient heat transfer.The surface area of the heating element comes into play in calculating how much heat it can release into the surrounding water.
An element’s design, including its length and diameter, results in a particular surface area directly influencing heat transfer capability:

• Surface area ( \( A_s \) ) = \( \pi D L \) \textrm{, where D and L are the diameter and length of the element respectively}

Maintaining a suitable design ensures that the electric heating element heats efficiently and safely.

Specific Heat

Specific heat refers to the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin).
For water, the specific heat, \( C_p \), is 4.186 kJ/kgK, reflecting how water requires a significant amount of energy to increase its temperature.This property is particularly important in calculating how long it takes to heat water to its boiling point.
The total energy required to heat water can be found using the formula:

• Energy ( \( Q \) ) = mass ( \( m \) ) x specific heat ( \( C_p \) ) x temperature change ( \( \Delta T \) )

A higher specific heat indicates that the substance can absorb more heat before its temperature rises significantly.
Understanding specific heat is critical in heating applications because it allows us to determine the time and power required to achieve desired temperature changes.