/*! 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 6 A motor supplies \(0.4\) hp to s... [FREE SOLUTION] | 91Ó°ÊÓ

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A motor supplies \(0.4\) hp to stir 5 kg of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water \(6{ }^{\circ} \mathrm{C}\) ? The heat required to heat the water is $$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6{ }^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$ This is actually supplied by friction work, so Friction work done \(=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}\) and this equals the work done by the motor. But Work done by motor in time \(t=(\) Power \()(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)\). Equating this to our previous value for the work done yields $$ t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min} $$

Short Answer

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It takes approximately 7 minutes.

Step by step solution

01

Calculate Heat Required to Heat Water

We are given that we need to heat 5 kg of water by 6°C. The heat required can be calculated using the formula: \( \Delta Q = mc\Delta T \). Here, \( m = 5000 \text{ g} \), \( c = 1 \text{ cal/g°C} \), and \( \Delta T = 6° \text{C} \). Substitute the values to find \( \Delta Q = 5000 \times 1 \times 6 = 30000 \text{ cal} = 30 \text{ kcal} \).
02

Convert Heat Energy to Joules

To convert the heat energy from calories to Joules, we use the conversion factor \( 1 \text{ cal} = 4.184 \text{ J} \). Thus, \( 30 \text{ kcal} = 30,000 \text{ cal} \times 4.184 \text{ J/cal} = 126,000 \text{ J} = 126 \text{ kJ} \).
03

Determine Work Done by Motor

The work done by the motor is equal to the heat energy calculated from the friction losses, which is \( 126 \text{ kJ} \). Since the power of the motor is given as \( 0.4 \text{ hp} \), we need to find time. Power in watts is \( 0.4 \text{ hp} \times 746 \text{ W/hp} = 298.4 \text{ W} \).
04

Calculate Time Required

We equate the work done by the motor to the heat energy from friction: \( \textrm{Work} = \text{Power} \times \text{time} \). So, \( 126,000 \text{ J} = 298.4 \text{ W} \times t \). Solving for \( t \) gives \( t = \frac{126,000}{298.4} \approx 422 \text{ seconds} \).
05

Convert Time to Minutes

The calculated time in seconds is approximately 422 seconds. To convert this to minutes: \( \frac{422 \text{ s}}{60 \text{ s/min}} \approx 7 \text{ minutes} \).

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Key Concepts

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

Work and Power
When we think about doing work in physics, it means that a force is acting over a distance. Power, on the other hand, is all about how quickly this work is done. It measures the rate of doing work or transferring energy.
For example, if you push a shopping cart, the work done is the force you apply multiplied by the distance the cart travels. Power would then measure how fast you moved the cart. If you push it quickly, you're using more power.
In mathematical terms, work is given by the equation \( ext{Work} = ext{Force} \times ext{Distance} \), and power is defined as \( ext{Power} = \frac{ ext{Work}}{ ext{Time}} \). In our exercise, the work done by the motor is the energy used to heat the water. The motor's power is 0.4 hp, which is a measure of how much work it can do in a certain time. To find out how long this work takes, we use the formula \( ext{Power} \times ext{Time} = ext{Work} \).
This relationship helps us understand that increasing the motor's power, or the time it runs, will increase the work done and consequently more energy transferred into heating the water.
Unit Conversion
Converting units is like translating languages; you're expressing the same quantity in different terms. This is crucial in physics because it helps us use formulas correctly and compare values accurately.
In the exercise, we convert power from horsepower to watts because the SI unit of power is the watt, not horsepower. We also convert thermal energy from calories to joules since energy calculations often use joules. One calorie is equal to 4.184 joules.
  • To convert horsepower to watts, we multiply by 746 (since 1 hp = 746 Watts).
  • To convert kcal to joules, we multiply by 4.184.
Unit conversions help ensure all parts of a calculation are consistent with each other. This consistency is important for accurate calculation and understanding of the problem. Whenever you're stuck with different units, remember it's about finding a common language for the problem.
Motor Efficiency
Motor efficiency refers to how well a motor converts electrical energy into mechanical energy. In an ideal world, all the energy used by a motor would go into doing the work we want. But reality includes some losses, like heat energy due to friction, which is evident in our exercise where energy from the motor is used to heat the water.
Efficiency is mathematically expressed as \( ext{Efficiency} = \frac{ ext{Useful Work Output}}{ ext{Total Energy Input}} \times 100\% \). In this case, all the work done by the motor is assumed to go into heating the water – an ideal scenario where the efficiency for this task is 100%.
However, typically, motors are not perfectly efficient due to energy lost as heat and to overcoming friction. The purpose of calculating efficiency is to understand how much energy is effectively used for the designed purpose versus how much is wasted. This helps in designing better motors and improving energy conservation in appliances and industrial applications.

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Most popular questions from this chapter

To determine the specific heat of an oil, an electrical heating coil is placed in a calorimeter with 380 g of the oil at \(10^{\circ} \mathrm{C}\). The coil consumes energy (and gives off heat) at the rate of \(84 \mathrm{~W}\). After \(3.0 \mathrm{~min}\), the oil temperature is \(40^{\circ} \mathrm{C}\). If the water equivalent of the calorimeter and coil is \(20 \mathrm{~g}\), what is the specific heat of the oil?

How much does the internal energy of \(5.0\) g of ice at precisely \(0{ }^{\circ} \mathrm{C}\) increase as it is changed to water at \(0{ }^{\circ} \mathrm{C}\) ? Neglect the change in volume. The heat needed to melt the ice is $$\Delta Q=m L_{f}=(5.0 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})=400 \mathrm{cal}$$ No external work is done by the ice as it melts and so \(\Delta W=0\). Therefore, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=(400 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})=1.7 \mathrm{~kJ} $$

Three kilomoles (6.00 kg) of hydrogen gas at S.T.P. expands isobarically to precisely twice its volume. (a) What is the final temperature of the gas? (b) What is the expansion work done by the gas? ( \(c\) ) By how much does the internal energy of the gas change? ( \(d\) ) How much heat enters the gas during the expansion? For \(\mathrm{H}_{2}, c_{v}=\) \(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assume the hydrogen will behave as an ideal gas. (a) From \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\) with \(P_{1}=P_{2}\), $$T_{2}=T_{1}\left(\frac{V_{2}}{V_{1}}\right)=(273 \mathrm{~K})(2.00)=546 \mathrm{~K}$$ (b) Because 1 kmol at S.T.P. occupies \(22.4 \mathrm{~m}^{3}\), we have \(V_{1}=\) \(67.2 \mathrm{~m}^{3}\). Then $$\Delta W=P \Delta V=P\left(V_{2}-V_{1}\right)=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(67.2 \mathrm{~m}^{3}\right)=6.8 \mathrm{MJ}$$ (c) To raise the temperature of this ideal gas by \(273 \mathrm{~K}\) at constant volume requires $$\Delta Q=c_{v} m \Delta T=(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(6.00 \mathrm{~kg})(273 \mathrm{~K})=16.4 \mathrm{MJ}$$ Because the volume is constant here, no work is done and \(\Delta Q\) equals the internal energy that must be added to the \(6.00 \mathrm{~kg}\) of \(\mathrm{H}_{2}\) to change its temperature from \(273 \mathrm{~K}\) to \(546 \mathrm{~K}\). Therefore, \(\Delta U=16.4 \mathrm{MJ} .\) ( \(d\) ) The system obeys the First Law during the process and so $$ \Delta Q=\Delta U+\Delta W=16.4 \mathrm{MJ}+6.8 \mathrm{MJ}=23.2 \mathrm{MJ} $$

Five moles of neon gas at \(2.00\) atm and \(27.0^{\circ} \mathrm{C}\) is adiabatically compressed to one-third its initial volume. Find the final pressure, final temperature, and external work done on the gas. For neon, \(\gamma\) \(=1.67, c_{v}=0.148 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and \(M=20.18 \mathrm{~kg} / \mathrm{kmol} .\)

A 70 -g metal block moving at \(200 \mathrm{~cm} / \mathrm{s}\) slides across a tabletop a distance of \(83 \mathrm{~cm}\) before it comes to rest. Assuming 75 percent of the thermal energy developed by friction goes into the block, how much does the temperature of the block rise? For the metal, \(c=\) \(0.106 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)

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