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(II) The 1.20 -kg head of a hammer has a speed of 7.5 \(\mathrm{m} / \mathrm{s}\) just before it strikes a nail (Fig. 29\()\) and is brought to rest. Estimate the temperature rise of a \(14-\mathrm{g}\) iron nail generated by 10 such hammer blows done in quick succession. Assume the nail absorbs all the energy.

Short Answer

Expert verified
The temperature rise of the nail is approximately 7.5°C.

Step by step solution

01

Calculate the kinetic energy of the hammer head

First, calculate the kinetic energy of the hammer head using the formula for kinetic energy: \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the hammer (1.20 kg) and \( v \) is the velocity (7.5 m/s). Substitute the given values: \[ KE = \frac{1}{2} \times 1.20 \times (7.5)^2 \]. Evaluate to find the kinetic energy.
02

Determine the total energy transferred

Since the hammer strikes the nail 10 times, the total energy transferred to the nail is 10 times the kinetic energy calculated in Step 1. Multiply the kinetic energy value from Step 1 by 10 to find the total energy absorbed by the nail.
03

Convert energy to temperature change

Use the specific heat capacity formula to find the temperature increase: \( Q = mc\Delta T \), where \( Q \) is the total energy, \( m \) is the mass of the nail (0.014 kg), \( c \) is the specific heat capacity of iron (450 J/kgâ‹…K), and \( \Delta T \) is the temperature change. Solve for \( \Delta T \): \[ \Delta T = \frac{Q}{mc} \].
04

Calculate \( \Delta T \)

Substitute the values for \( Q \), \( m \), and \( c \) into \( \Delta T = \frac{Q}{mc} \) to calculate the temperature change in the nail.

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Key Concepts

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

Specific Heat Capacity
Specific Heat Capacity is an essential concept in physics that describes how much energy is required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. It's a property that varies from material to material. In our problem, we are looking at iron, which has a specific heat capacity denoted by \( c \). For iron, this value is \( 450 \text{ J/kgâ‹…K} \). This means that it takes 450 joules of energy to increase the temperature of 1 kilogram of iron by 1 degree Kelvin or Celsius.

Understanding specific heat capacity helps you predict how different materials will react when energy is applied to them, like the iron nail in this exercise. By knowing the specific heat capacity, you can use the energy involved to calculate the resulting temperature change. This is done with the formula \( Q = mc\Delta T \), where \( Q \) is the energy transferred, \( m \) is the mass of the object (in kilograms), and \( \Delta T \) is the temperature change.

For the nail in our exercise, the specific heat capacity allows us to relate the energy from the hammer blows to a rise in temperature. This understanding is crucial in analyzing how objects heat up and how thermal energy transfers through various materials.
Kinetic Energy Formula
The Kinetic Energy Formula is used to calculate the energy of a moving object. The formula is \( KE = \frac{1}{2} mv^2 \), where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is the object's velocity. This equation helps you understand how the movement of an object can be described as energy. The mass and the square of the velocity determine how much kinetic energy the object has.

In our exercise, the hammer head has a mass of 1.20 kg and a speed of 7.5 m/s. By substituting these values into the kinetic energy formula, we can find how much energy the hammer has just before impacting the nail. This is crucial because this energy, when transferred, affects the temperature change of the nail.

One key point to remember is that kinetic energy depends highly on the velocity. If the velocity is doubled, the kinetic energy increases by four times. This quadratic relationship shows why speed is a critical factor in kinetic energy calculations. For our scenario, since 10 hammer blows occur, this energy is multiplied, meaning the total kinetic energy transferred to the nail is 10 times the energy calculated from a single blow.
Energy Transfer Calculation
Energy Transfer Calculation involves determining how energy moves from one object to another and what effect it has. In the exercise at hand, we need to track how the kinetic energy of the hammer is converted into thermal energy in the nail. This means using the energy calculated from the kinetic formula and multiplying it by the number of hammer strikes.

Once we have the total kinetic energy transferred (calculated in the previous steps), we use it in the equation \( Q = mc\Delta T \), where \( Q \) is the total energy transferred to the nail. \( m \) is the mass of the nail (0.014 kg), \( c \) is the specific heat capacity of iron (450 J/kgâ‹…K), and \( \Delta T \) is the temperature change we're solving for.

This calculation gives us the temperature increase of the nail as a result of the kinetic energy from the hammer being absorbed. It's fascinating to see how mathematically, all these factors intertwine to show how energy conversion results in physical changes, such as a temperature increase in this scenario. Calculating these changes is critical in understanding the broader concepts of energy conservation in physics.

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Most popular questions from this chapter

(III) A 1.00 -mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of \(420 \mathrm{~K}\) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are \(720 \mathrm{~K}\) and 1.60 atm. Determine \((a)\) the change in internal energy, (b) the work done by the gas, and (c) the heat added to the gas. (Assume five active degrees of freedom.)

(II) A British thermal unit (Btu) is a unit of heat in the British system of units. One Btu is defined as the heat needed to raise 1 lb of water by 1 \(\mathrm{F}^{\circ} .\) Show that $$1 \mathrm{B} \mathrm{tu}=0.252 \mathrm{kcal}=1056 \mathrm{J}$$

A marathon runner has an average metabolism rate of about 950 \(\mathrm{kcal} / \mathrm{h}\) during a race. If the runner has a mass of 55 \(\mathrm{kg}\) , estimate how much water she would lose to evaporation from the skin for a race that lasts 2.2 \(\mathrm{h}\) .

(II) \((a)\) How much energy is required to bring a \(1.0-\mathrm{L}\) pot of water at \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C} ?(b)\) For how long could this amount of energy run a \(100-\mathrm{W}\) lightbulb?

(II) Suppose 2.60 \(\mathrm{mol}\) of \(\mathrm{an}\) ideal gas of volume \(V_{1}=3.50 \mathrm{m}^{3}\) at \(T_{1}=290 \mathrm{K}\) is allowed to expand isothermally to \(V_{2}=7.00 \mathrm{m}^{3}\) at \(T_{2}=290 \mathrm{K}\) . Determine (a) the work done by the gas, \((b)\) the heat added to the gas, and \((c)\) the change in internal energy of the gas.

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