/*! 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 147 A 2-kW electric resistance heate... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for \(10 \mathrm{~min}\). During the process, \(500 \mathrm{~kJ}\) of heat is lost from the water. The temperature rise of water is (a) \(5.6^{\circ} \mathrm{C}\) (b) \(9.6^{\circ} \mathrm{C}\) (c) \(13.6^{\circ} \mathrm{C}\) (d) \(23.3^{\circ} \mathrm{C}\) (e) \(42.5^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (a) 5.6°C

Step by step solution

01

Convert time to seconds.

Since the heater is turned on for 10 minutes, we need to convert the time to seconds for further calculations. 1 minute equals 60 seconds, so 10 minutes equals $$10 \times 60 = 600 \mathrm{~s}$$.
02

Calculate the energy provided by the heater.

The energy provided by the heater can be calculated using the formula $$E = Pt$$, where E is the energy, P is the power of the heater, and t is the time the heater is on. We have $$P = 2 \mathrm{~kW} = 2000 \mathrm{~W}$$ and $$t = 600 \mathrm{~s}$$. Therefore, the energy provided by the heater is $$E = 2000 \mathrm{~W} \times 600 \mathrm{~s} = 1200 \mathrm{~kJ}$$.
03

Calculate the net energy absorbed by the water.

We're given that 500 kJ of heat is lost during the process. So, the net energy absorbed by the water can be found by subtracting the heat loss from the energy provided by the heater: $$E_{\mathrm{net}} = 1200 \mathrm{~kJ} - 500 \mathrm{~kJ} = 700 \mathrm{~kJ}$$.
04

Calculate the temperature rise of water.

The formula to calculate the temperature change with the energy absorbed and the mass of water is $$\Delta T = \frac{E_{\mathrm{net}}}{mc_{\mathrm{p}}}$$, where \(\Delta T\) is the temperature change, E\(_{\mathrm{net}}\) is the net energy absorbed, m is the mass of the water, and \(c_{\mathrm{p}}\) is the specific heat capacity of the water. For water, $$c_{\mathrm{p}} = 4.18 \mathrm{~kJ/kg}^{\circ} \mathrm{C}$$ and $$m = 30 \mathrm{~kg}$$. Therefore, the temperature rise of water is $$\Delta T = \frac{700 \mathrm{~kJ}}{30 \mathrm{~kg} \times 4.18 \mathrm{~kJ/kg}^{\circ} \mathrm{C}} = 5.58^{\circ} \mathrm{C}$$. The temperature rise of water is approximately \(5.6^{\circ} \mathrm{C}\), which corresponds to the answer (a).

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Key Concepts

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

Heat Energy Transfer
Understanding how heat energy moves from one object to another is fundamental in physics and is crucial for solving problems like the one presented. Heat energy transfer can occur through three primary methods: conduction, convection, and radiation. In our exercise, we focus on the concept of heat transfer through conduction, as the electric resistance heater directly interacts with the water.

During the operation of the heater, it converts electrical energy into heat energy due to its resistance. This heat is then transferred to the water. However, not all of the heat is retained; some of it is inevitably lost to the surroundings. That's where the importance of calculating the net energy absorbed by the water comes into play. The exercise highlights that 500 kJ of heat escapes, which must be accounted for to find the actual temperature increase of the water. When solving similar problems, be mindful of the surrounding environment, which can affect the net heat transfer.
Specific Heat Capacity
The term 'specific heat capacity' (\(c_p\)) might seem daunting, but it's simply a property that indicates how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. It's like a 'thermal bank account' for a material, showing the heat 'savings' required for a temperature 'transaction.'

The correct understanding of specific heat capacity is vital for calculating temperature changes, as seen in our exercise. Water, with a specific heat capacity of 4.18 kJ/kg°C, can 'hold' a considerable amount of heat before its temperature changes significantly. This inherent trait of water, or any other substance's specific heat capacity, must be factored into our temperature rise calculation. If unsure about the specific heat capacity values, you can generally find them in physics textbooks or reliable scientific databases.
Electric Resistance Heater
An electric resistance heater is a common device in various heating applications, both practical and experimental. By passing an electric current through a resistive material, it utilizes the principle of Joule heating; the resistance to the current flow generates heat. This is often seen in household items like toasters or electric stoves.

In our exercise, the power rating of 2 kW indicates the rate at which the heater uses energy to produce heat. Tying it into our earlier concepts, the heat energy transfer from this heater to the water determines the rise in temperature. It's crucial to convert the power to a common unit of energy (joules or kilojoules) and to account for time to find out how much energy is actually transferred. This is a good example of practical thermodynamics and gives a clear view of energy transformation in everyday appliances.

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Most popular questions from this chapter

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

How does heat conduction differ from convection?

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

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

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