A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to \(9000 \mathrm{~kJ} / \mathrm{h}\), it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in \(\mathrm{kW}\).

Short Answer

Expert verified
Answer: The power rating of the heater is 9 kW.

Step by step solution

01

Determine the energy loss in the room per hour.

Our given energy loss per hour is 9000 kJ/h.
02

Convert the energy loss into power rating.

Now, we need to find the power rating in kW. We can do this by dividing the energy loss with time in hours and converting the answer from kJ to kW. 1 kJ = 0.001 kW (since 1 kW = 1000 kJ) We know that power (P) is given by: \(P = \frac{Energy}{Time}\)
03

Calculate the power rating of the heater.

Now, we plug in the energy loss and time into the power formula: \(P = \frac{9000 \mathrm{~kJ}}{1 \mathrm{~h}}\) \(P = 9000 \mathrm{~kJ/h}\) . Now, we convert the power from kJ/h to kW: \(P = 9000 * 0.001 \mathrm{~kW}\) \(P = 9 \mathrm{~kW}\) Hence, the power rating of the heater is 9 kW.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Loss
Heat loss refers to the transfer of thermal energy from a warmer space to a cooler space. In the context of a heated room, heat loss occurs through various pathways like windows, walls, floors and even through the air exchange with the outside. This process is driven by the temperature difference between the inside of the room and the colder external environment.

Understanding heat loss is crucial when calculating the heating requirements for maintaining a constant temperature within a space. The amount of heat loss must be countered with an equal quantity of heat generation to achieve thermal equilibrium. In our exercise, the room's heat loss was given as 9000 kJ/h, indicating the amount of energy being lost to the surroundings every hour.
Energy Conversion
Energy conversion is the process of changing one form of energy to another. In the case of resistance heaters, electrical energy is converted into thermal energy to provide heating. This conversion is rarely 100% efficient due to losses like those from the material's resistance.

The efficiency of the conversion process affects how much electrical power is needed to sustain the desired level of thermal energy within a room. Since our exercise deals with a scenario where the temperature remains constant, it is implied that the electrical power supplied to the heater is being fully converted into heat to balance the heat lost.
Power Formula
The power formula, expressed as Power (P) equals Energy (E) divided by Time (T), provides a way to calculate the rate at which energy is used or generated. In terms of units, Power is measured in watts (W) which is equivalent to joules per second (J/s), and in this case, we are also considering the kilowatt (1 kW = 1000 W).

Applying the power formula to our problem, we determine the power rating of the heater by dividing the energy loss (expressed in kilojoules) by the time (in hours) and then converting the result into kilowatts. This calculation tells us the steady power output that the heater must provide to maintain constant temperature despite the heat loss.
Thermal Energy
Thermal energy is the internal energy of an object due to the movement of its atoms and molecules. It's commonly manifested as heat. The amount of thermal energy in a substance often depends on its temperature and mass.

In resistance heating, an electric current passes through a resistant material, causing it to heat up due to the friction of moving electrons. This heat is then transferred to the surrounding environment, like our room in the exercise. To keep the room's temperature steady, the thermal energy supplied by the heater must match the thermal energy lost, which we have quantified as heat loss.
Resistance Heating
Resistance heating is a method of converting electrical energy into heat energy by passing an electric current through a resistor. This is the principle behind devices like baseboard heaters. The electric current meets resistance as it passes through the heating element, which causes the element to heat up.

Factors that affect resistance heating efficiency include the material of the resistor, its temperature, and the surrounding environment. These must be considered when designing heating systems. In our textbook problem, the continuous operation of the baseboard resistance heater with a power rating of 9 kW suggests it has been well-matched to the room's heat loss rate to maintain a constant temperature.

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Most popular questions from this chapter

The critical heat flux (CHF) is a thermal limit at which a boiling crisis occurs whereby an abrupt rise in temperature causes overheating on fuel rod surface that leads to damage. A cylindrical fuel rod of \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of \(150 \mathrm{MW} / \mathrm{m}^{3}\). The average temperature of the cooling water, sufficiently far from the fuel rod, is \(80^{\circ} \mathrm{C}\). The operating pressure of the cooling water is such that the surface temperature of the fuel rod must be kept below \(300^{\circ} \mathrm{C}\) to avoid the cooling water from reaching the critical heat flux. Determine the necessary convection heat transfer coefficient to avoid the critical heat flux from occurring.

Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \((0.75,1\), and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, \(\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]\) for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

A cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

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