Question

A \(4-m \times 5-m \times 6-m\) room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 5 to \(25^{\circ} \mathrm{C}\) within 11 min. Assuming no heat losses from the room and an atmospheric pressure of \(100 \mathrm{kPa}\), determine the required power of the resistance heater. Assume constant specific heats at room temperature.

Short answer

Answer: The required power of the resistance heater is approximately 431.515 kW.

Step-by-step solution

Calculate the volume of the room

To find the volume (V) of the room, multiply its dimensions: \(V = length \cdot width \cdot height\). So, \(V = 4\,\text{m} \times 5\,\text{m} \times 6\,\text{m} = 120\,\text{m}^3\)

Calculate the mass of the air

Assume the air behaves as an ideal gas. The ideal gas law states \(PV = mRT\), where P is pressure, V is volume, m is mass, R is the specific gas constant, and T is temperature. We are given \(P = 100\,\text{kPa}\) and \(T_1 = 5\,^{\circ}\text{C} = 278\,\text{K}\) (converting to Kelvin). Also, for air, \(R = 0.287\,\text{kPa}\cdot \text{m}^3/\text{kg}\cdot\text{K}\). Rearrange the equation to solve for mass (m): \(m = \frac{PV}{RT}\) Plug in the values: \(m = \frac{100\,\text{kPa} \times 120\,\text{m}^3}{0.287\,\text{kPa}\cdot\text{m}^3/\text{kg}\cdot\text{K}\times 278\,\text{K}} = 14180.84\,\text{kg}\)

Calculate the heat needed

Use the equation \(Q = mc\Delta T\), where Q is the heat, m is mass, c is specific heat, and \(\Delta T\) is the change in temperature. The specific heat of air at room temperature is approximately \(c = 1.006\,\text{kJ/kg}\cdot\text{K}\), and \(\Delta T = T_2 - T_1 = 25\,^{\circ}\text{C} - 5\,^{\circ}\text{C} = 20\,^{\circ}\text{C}\). Plug in the values: \(Q = 14180.84\,\text{kg} \times 1.006\,\text{kJ/kg}\cdot\text{K} \times 20\,^{\circ}\text{C} = 284800\,\text{kJ}\)

Calculate the required power

Use the equation \(P_{required} = \frac{Q}{t}\), where P is power, Q is the heat, and t is the time taken. The given time interval is 11 minutes, which should be converted to seconds: \(t = 11\,\text{min} \times 60\,\text{s/min} = 660\,\text{s}\). Plug in the values: \(P_{required} = \frac{284800\,\text{kJ}}{660\,\text{s}} = 431.515\,\text{kW}\) Therefore, the required power of the resistance heater is approximately \(431.515\,\text{kW}\).

Key concepts

Resistance Heater Power Calculation

Understanding the power calculation of a resistance heater is essential when determining the capability of a heating system to increase the temperature of a space within a specified timeframe. In the given exercise, this entails calculating the power required to raise the temperature of air in a room from 5 to 25 degrees Celsius in only 11 minutes.

To calculate the required heater power, first, we need to ascertain the amount of energy, or heat, needed to achieve the desired temperature increase. This is where the concept of the specific heat capacity comes in, which will be discussed in the next section. Once we have the total heat required, we divide it by the time (in seconds) to get the power in kilowatts.

It's important to note that power, denoted by 'P', is the rate of energy transfer over time. The formula used is: \[P_{\text{required}} = \frac{Q}{t}\], where 'Q' is the heat energy in kilojoules (kJ) and 't' is the time in seconds. This gives a clear indication of the resistance heater's power requirement to fulfill the heating need, thus making it possible to select an appropriate heater to match the output.

Specific Heat Capacity

Specific heat capacity is a key concept when analyzing heat transfer within materials, including gases like air. It represents the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. To understand this better, consider it as a measure of a material's thermal inertia; a higher specific heat capacity means more energy is needed to change its temperature.

For air at room temperature, the specific heat capacity is approximately 1.006 kJ/kg°C. In the exercise, this value is critical for calculating the total heat energy (Q) needed for the air in the room to reach the desired temperature. The equation, \[Q = mc\Delta T\], where 'm' is the mass of air, 'c' is the specific heat capacity, and '\Delta T' is the temperature difference, is used to calculate the total heat required for the room's air volume. By understanding specific heat capacity, students can grasp why different substances heat up at different rates and why some materials are better for heat storage than others.

Heat Transfer

Heat transfer encompasses the movement of thermal energy from one place to another and is central to the process of heating a room with a resistance heater. There are three modes of heat transfer: conduction, convection, and radiation. In our exercise, the resistance heater transfers heat to the room's air primarily through convection, where the heated air becomes less dense and rises, subsequently being replaced by cooler air that gets heated in turn.

It is crucial to appreciate how heat transfer operates, especially in a controlled environment like the room in the exercise. Here, we make a simplifying assumption that there are no heat losses, meaning all the energy from the heater is effectively used to raise the room temperature. The efficiency and speed of heat transfer are influenced by the specific heat capacity and the mass of air, reinforcing why these concepts are intertwined with the calculation of the heater's power requirement. An in-depth understanding of heat transfer not only aids in solving textbook problems but is also foundational knowledge for a multitude of real-life applications, from HVAC system design to thermal management in electronics.