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Chapter 1: Problem 146

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)

Short Answer

Expert verified
a) 1000 kJ b) 5000 kJ c) 6000 kJ d) 7000 kJ Answer: c) 6000 kJ

Step by step solution

01

Convert the time to seconds

To convert the time from minutes to seconds, we should multiply the time in minutes (50) by 60 seconds, as there are 60 seconds in 1 minute: Time (seconds) = 50 minutes x 60 seconds/minute = 3000 seconds
02

Convert the power to Watts

The power of the heater is given in kilowatts (kW), and we need to convert it to watts (W). To do this, we should multiply the power in kilowatts (2) by 1000 watts, as there are 1000 watts in 1 kilowatt: Power (Watts) = 2 kW x 1000 watts/kW = 2000 watts
03

Calculate the energy transferred to the room

Now, we can use the formula for calculating the energy transferred to the room: Energy (Joules) = Power (Watts) x Time (Seconds) Energy (Joules) = 2000 watts x 3000 seconds = 6,000,000 J As the energy is given in kJ (kilojoules) in the exercise options, we should convert the energy from Joules (6,000,000 J) to kilojoules by dividing it by 1000: Energy (kJ) = 6,000,000 J 梅 1000 = 6000 kJ
04

Match the result with given options

The calculated energy transferred to the room is 6000 kJ. Comparing it to the given options, we find that it corresponds to the option (c) 6000 kJ. So, the correct answer is (c) 6000 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. This is achieved through a process known as resistive heating. The heater contains a resistive element, typically made of materials such as nichrome. When electric current passes through this element, it encounters resistance which converts electrical energy into heat.
The generated heat is then transferred to the surrounding area, such as a room, to raise its temperature.
  • Electric resistance heaters are simple and reliable, often used as space heaters in homes and offices.
  • They are easy to control, allowing for quick adjustment of temperature output.
  • However, they can be less efficient overall compared to other heating methods because they consume substantial electrical energy.
Understanding how electric resistance heaters work is fundamental to grasp the concept of energy conversion and power calculation.
Energy Conversion
Energy conversion in an electric resistance heater involves transforming electrical energy into thermal energy. This transformation follows the law of conservation of energy, which states that energy cannot be created nor destroyed, only changed from one form to another.
When we turn on an electric resistance heater, the electrical energy flowing through the resistive wire is converted to thermal energy due to the resistance. This thermal energy is then used to heat the room.
  • The energy conversion efficiency in an electric resistance heater is determined primarily by the resistance of the wire and the power of the electrical source.
  • The higher the power and resistance, the greater the amount of thermal energy produced.
Such heaters are effective for direct heating applications because nearly all the electrical energy is converted directly into heat. Hence, understanding energy conversion helps explain how heaters warm up spaces effectively and calculate the amount of energy delivered to a room.
Power Calculation
The power calculation is crucial for determining how much energy is transferred in a given time by a device like an electric resistance heater. Power is the rate at which energy is used or transferred. It is typically measured in watts (W) or kilowatts (kW).
In this exercise, the power of the heater is 2 kW, which means it uses 2000 Joules of energy per second. To find the total energy transferred over a period, you must multiply the power by the time the heater runs.
A step-by-step power calculation is as follows:
  • First, convert time from minutes to seconds because power is typically given in watts, meaning Joules per second.
  • Multiply power (in watts) by time (in seconds) to get total energy in Joules.
  • Convert Joules to kilojoules by dividing by 1000, as energy is often expressed in kilojoules for convenience.
For the given problem, the calculation shows that during 50 minutes, or 3000 seconds, the 2 kW heater transfers 6000 kJ of energy to the room. Power calculations are vital in assessing and optimizing energy usage in heating applications.

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Most popular questions from this chapter

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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.

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

An engine block with a surface area measured to be \(0.95 \mathrm{~m}^{2}\) generates a power output of \(50 \mathrm{~kW}\) with a net engine efficiency of \(35 \%\). The engine block operates inside a compartment at \(157^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If convection is the only heat transfer mechanism occurring, determine the engine block surface temperature.

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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