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Chapter 18: Problem 35

A \(0.250\) -hp drill causes a dull \(50.0-\mathrm{g}\) steel bit to heat up rather than to deepen a hole in a block of hard wood. Assuming that \(75.0\) percent of the friction-loss energy causes heating of the bit, by what amount will its temperature change in \(20.0 \mathrm{~s}\) ? For steel, \(c=450 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The temperature change is approximately \(124.3^{\circ} \text{C}\).

Step by step solution

01

Convert Horsepower to Watts

The drill's power is given in horsepower (hp), which we need to convert to watts (W) to use it in further calculations. We know that:\[ 1 ext{ hp} = 746 ext{ W} \]Therefore, a \(0.250\) hp drill is:\[ 0.250 imes 746 = 186.5 ext{ W} \]
02

Calculate Total Energy Provided

Next, we calculate the total energy provided by the drill in \(20.0\) seconds. Using the formula for power, \(P = \frac{E}{t}\), where \(E\) is energy and \(t\) is time, we find that:\[ E = P \times t = 186.5 ext{ W} \times 20.0 ext{ s} = 3730 ext{ J} \]
03

Calculate Energy Causing Heating

Since only \(75\%\) of the energy goes into heating the bit, we calculate the fraction of total energy used for heating:\[ E_{heat} = 0.75 \times 3730 = 2797.5 ext{ J} \]
04

Convert Mass to Kilograms

The mass of the steel bit is given as \(50.0\) grams. We need to convert this to kilograms for consistency with the specific heat capacity units:\[ 50.0 ext{ g} = 0.0500 ext{ kg} \]
05

Calculate Temperature Change

Using the formula that relates energy to temperature change for a specific material, \(Q = mc\Delta T\), where \(Q\) is heat energy, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the temperature change:Rearranging for \(\Delta T\), we have:\[ \Delta T = \frac{Q}{mc} = \frac{2797.5}{0.0500 \times 450} \approx 124.3^{\circ} \text{C} \]
06

Final Discussion

The temperature of the steel bit will increase by approximately \(124.3^{\circ} \text{C}\) over the \(20.0\) seconds due to the frictional heating from the drill.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept within thermodynamics. It involves the movement of thermal energy from one object or substance to another. In the case of the drill and the steel bit, heat is transferred from the drill's action (energy conversion) to the steel bit.
  • Heat transfer occurs due to a temperature difference between two objects or within different parts of a single object.
  • The three primary modes of heat transfer are conduction, convection, and radiation.
  • In this scenario, conduction is the mode of heat transfer, as heat is conducted from the point of friction between the drill and the bit.
Understanding these modes helps explain how heat disperses in various processes and why materials get hot with frictional work, as seen in the drill bit example.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius.
  • It represents the thermal inertia of a material, i.e., its resistance to temperature change.
  • Materials with a high specific heat capacity require more energy to change temperature.
  • In the steel bit problem, knowing the specific heat capacity of steel (\(450 \, \mathrm{J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\)) is crucial to calculate how much the temperature will rise with given energy input.
Specific heat capacity allows engineers and scientists to predict how materials respond to energy changes, an essential factor in material design and safety.
Energy Conversion
Energy conversion involves transforming energy from one form to another, such as chemical to mechanical or mechanical to thermal energy. In the problem at hand, the drill converts mechanical energy into thermal energy:
  • The drill operates using electrical or chemical energy which it converts to mechanical energy.
  • This mechanical energy experiences a loss due to friction between the drill bit and the workpiece, converted into heat.
  • The efficiency of energy conversion determines how much of the initial energy ends up as useful work versus heat, as demonstrated by the 75% efficiency figure given.
Understanding energy conversion helps in optimizing processes for better energy efficiency and lesser waste, which are vital in energy management and environmental conservation.
Temperature Change
Temperature change quantifies how much an object's temperature increases or decreases due to energy transfer.
  • When heat energy is added to or removed from a material, its temperature changes reflecting the quantity of heat exchanged.
  • The mathematical relationship is given by the equation: \(\Delta T = \frac{Q}{mc}\)
  • For the steel bit, we calculated an increase of approximately \(124.3^{\circ} \mathrm{C}\) as a result of energy input from friction.
Calculating temperature change is critical in thermodynamics for designing thermal systems and ensuring material stability and performance during operation.

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Most popular questions from this chapter

Determine the result when \(10 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is passed into a mixture of \(400 \mathrm{~g}\) of water and \(100 \mathrm{~g}\) of ice at exactly \(0{ }^{\circ} \mathrm{C}\) in a calorimeter which behaves thermally as if it were equivalent to \(50 \mathrm{~g}\) of water.

A \(20-\mathrm{g}\) piece of aluminum \(\left(c=0.21 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(90^{\circ} \mathrm{C}\) is dropped into a cavity in a large block of ice at \(0{ }^{\circ} \mathrm{C}\). How much ice does the aluminum melt? (Heat change of Al as it cools to \(0{ }^{\circ} \mathrm{C}\) ) \(+\) (Heat change of mass \(m\) of ice melted) \(=0\) $$ \begin{array}{r} (m c \Delta T)_{\mathrm{Al}}+\left(L_{f} m\right)_{\mathrm{ice}}=0 \\ (20 \mathrm{~g})\left(0.21 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(0{ }^{\circ} \mathrm{C}-90^{\circ} \mathrm{C}\right)+(80 \mathrm{cal} / \mathrm{g}) m=0 \end{array} $$ from which \(m=4.7 \mathrm{~g}\) is the quantity of ice melted.

Furnace oil has a heat of combustion of \(44 \mathrm{MJ} / \mathrm{kg}\). Assuming that 70 percent of the heat is useful, how many kilograms of oil are required to raise the temperature of \(2000 \mathrm{~kg}\) of water from \(20{ }^{\circ} \mathrm{C}\) to \(99{ }^{\circ} \mathrm{C}\) ?

Determine the temperature that results when \(1.0 \mathrm{~kg}\) of ice at exactly \(0{ }^{\circ} \mathrm{C}\) is mixed with \(9.0 \mathrm{~kg}\) of water at \(50^{\circ} \mathrm{C}\) and no heat is lost.

Two identical metal plates (mass \(=m\), specific heat \(=c\) ) have different temperatures; one is at \(20{ }^{\circ} \mathrm{C}\), and the other is at \(90^{\circ} \mathrm{C}\). They are placed in good thermal contact. What is their final temperature? Because the plates are identical, we would guess the final temperature to be midway between \(20^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\), namely \(55^{\circ} \mathrm{C}\). This is correct, but let us show it mathematically. From the law of conservation of energy, the heat lost by one plate must equal the heat gained by the other. Thus, the total heat change of the system is zero. In equation form, (Heat change of hot plate) \(+(\) Heat change of cold plate \()=0\) $$ m c(\Delta T)_{\mathrm{hot}}+m c(\Delta T)_{\mathrm{cold}}=0 $$ which is short-hand for \(m_{\text {hot }} c_{\text {hot }} \Delta T_{\text {hot }}+m_{\text {cold }} c_{\text {cold }} \Delta T_{\text {cold }}=0\). Be careful about \(\Delta T\) : It is the final temperature (which we denote by \(T_{f}\) in this case) minus the initial temperature. The above equation thus becomes $$ m c\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+m c\left(T_{f}-20{ }^{\circ} \mathrm{C}\right)=0 $$ After canceling \(m c\) from each term, solve the equation and find \(T_{f}=55^{\circ} \mathrm{C}\), the expected answer.
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