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A \(2-k W\) electric resistance heater in a room is turned on and kept on for 50 min. The amount of energy transferred to the room by the heater is \((a) 2 \mathrm{kJ}\) \((b) 100 \mathrm{kJ}\) \((c) 3000 \mathrm{kJ}\) \((d) 6000 \mathrm{kJ}\) \((e) 12,000 \mathrm{kJ}\)

Short Answer

Expert verified
Answer: (d) 6000 kJ

Step by step solution

01

Convert power and time to appropriate units

The power of the heater is given as 2 kW. We need to convert this value into watts (W): \(\text{Power (P)} = 2 \text{kW} \times 1000 \text{W/kW} = 2000 \text{W}\) The time is given as 50 minutes. We need to convert this value into seconds: \(\text{Time (t)} = 50 \text{ min} \times 60 \text{s/min} = 3000 \text{s}\)
02

Calculate the energy transferred

Now, using the formula for energy (E = P × t), we can calculate the total energy transferred to the room: \(\text{Energy (E)} = \text{Power (P)} \times \text{Time (t)}\) \(E = 2000 \text{W} \times 3000 \text{s} = 6000000 \text{J}\)
03

Convert energy to kilojoules

To compare with the options given in the exercise, we need to convert the energy to kilojoules (kJ): \(\text{Energy (E)} = 6000000 \text{J} \times \dfrac{1 \text{kJ}}{1000 \text{J}} = 6000 \text{kJ}\) The amount of energy transferred to the room by the heater is 6000 kJ, which corresponds to option (d). So, the correct answer is \((d) 6000 \mathrm{kJ}\).

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Key Concepts

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

Electric Resistance Heater
An electric resistance heater is a device that converts electrical energy into heat energy through the process of resistance heating. When an electrical current passes through a resistive material, it encounters opposition to flow, known as resistance. This resistance causes electric energy to be converted into heat, which is then transferred to the surroundings.

For instance, in a 2 kW heater, the '2 kW' refers to the power rating of the heater, indicating the rate at which it can convert electrical energy into heat energy. The higher the power rating, the more heat can be generated and transferred to the environment in a given amount of time. In our exercise, the heater operates at 2 kW, providing a substantial amount of heat to warm the room.
Power Unit Conversion
Power unit conversion is essential when calculating energy transfer because power units often come in different scales. In physics and engineering, the standard unit of power is the watt (W), which is defined as one joule per second. However, given that power can range from very small to very large values, kilowatts (kW) and megawatts (MW) are also commonly used, where 1 kW equals 1000 W and 1 MW equals 1,000,000 W.

To solve our exercise correctly, we first convert the power of the electric resistance heater from kilowatts to watts to match the standard unit for further calculations:
  • 2 kW = 2 × 1000 W/kW = 2000 W
Time Unit Conversion
Time unit conversion is another critical step in energy calculation. Engineers and scientists commonly measure time in seconds when calculating energy transfer, as the standard unit of energy, the joule, is defined in terms of seconds. However, in everyday situations, time might be given in minutes, hours, or even longer periods.

For accurate energy calculations, converting to seconds is needed because it directly relates to the definition of power (watts). In the exercise:
  • 50 minutes = 50 × 60 seconds/minute = 3000 seconds
Without converting minutes to seconds, the calculation of energy transferred through an electric resistance heater would not be accurate.
Energy Formula Application
The energy formula application is pivotal to understanding how much energy is transferred from one system to another. Energy transferred (E) can be calculated using the formula:
\[ E = P \times t \]
where 'E' is the energy transferred in joules (J), 'P' is the power in watts (W), and 't' is the time in seconds (s).

In our exercise, after converting the heater’s power to watts and the operating time to seconds, we apply the energy formula to find the total energy transferred to the room:
  • \( E = 2000 \text{W} \times 3000 \text{s} = 6000000 \text{J} \)
By using these conversions and the energy formula, we ensure the calculation is accurate and meaningful, leading to the correct answer for the energy transferred.

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Most popular questions from this chapter

A 75 -hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high efficiency motor that has an efficiency of 95.4 percent. The motor operates 4368 hours a year at a load factor of 0.75 Taking the cost of electricity to be \(\$ 0.12 / \mathrm{kWh}\), determine the amount of energy and money saved as a result of installing the high-efficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are \(\$ 5449\) and \(\$ 5520,\) respectively.

A 50 -cm-long, 0.2 -cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be \(130^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be 4.1 \(\mathrm{kW}\). Then the heat transfer coefficient is \((a) 43,500 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((b) 137 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((c) 68,330 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((d) 10,038 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((e) 37,540 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\)

At winter design conditions, a house is projected to lose heat at a rate of \(60,000 \mathrm{Btu} / \mathrm{h}\). The internal heat gain from people, lights, and appliances is estimated to be \(6000 \mathrm{Btu} / \mathrm{h}\) If this house is to be heated by electric resistance heaters, determine the required rated power of these heaters in \(\mathrm{kW}\) to maintain the house at constant temperature.

On a hot summer day, the air in a well-sealed room is circulated by a 0.50 -hp fan driven by a 65 percent efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan.) The rate of energy supply from the fanmotor assembly to the room is \((a) 0.769 \mathrm{kJ} / \mathrm{s}\) \((b) 0.325 \mathrm{kJ} / \mathrm{s}\) \((c) 0.574 \mathrm{kJ} / \mathrm{s}\) \((d) 0.373 \mathrm{kJ} / \mathrm{s}\) \((e) 0.242 \mathrm{kJ} / \mathrm{s}\)

A construction crane lifts a prestressed concrete beam weighing 3 short tons from the ground to the top of piers that are \(36 \mathrm{ft}\) above the ground. Determine the amount of work done considering ( \(a\) ) the beam and ( \(b\) ) the crane as the system. Express your answers in both lbf-ft and Btu.

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