/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A room is heated by a baseboard ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

In a manufacturing plant, AISI 1010 carbon steel strips \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}\right)\) of \(2 \mathrm{~mm}\) thick and \(3 \mathrm{~cm}\) wide are conveyed into a chamber at a constant speed to be cooled from \(527^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). Determine the speed of a steel strip being conveyed inside the chamber, if the rate of heat being removed from a steel strip inside the chamber is \(100 \mathrm{~kW}\).

Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of \(150 \mathrm{~m}^{2}\) and an average height of \(3 \mathrm{~m}\) at \(1000 \mathrm{~m}\) elevation, where the standard atmospheric pressure is \(89.6 \mathrm{kPa}\). The house is maintained at a temperature of \(22^{\circ} \mathrm{C}\), and the infiltration losses are estimated to amount to \(0.7 \mathrm{ACH}\). Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is \(5^{\circ} \mathrm{C}\). Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $$\$ 0.082 / \mathrm{kWh}$$

What is the physical mechanism of heat conduction in a solid, a liquid, and a gas?

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

Consider a house with a floor space of \(200 \mathrm{~m}^{2}\) and an average height of \(3 \mathrm{~m}\) at sea level, where the standard atmospheric pressure is \(101.3 \mathrm{kPa}\). Initially the house is at a uniform temperature of \(10^{\circ} \mathrm{C}\). Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of \(22^{\circ} \mathrm{C}\). Determine how much heat is absorbed by the air assuming some air escapes through the cracks as the heated air in the house expands at constant pressure. Also, determine the cost of this heat if the unit cost of electricity in that area is $$\$ 0.075 / \mathrm{kWh}$$.

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