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Chapter 14: Problem 27

\bullet A 15.0 g bullet traveling horizontally at 865 \(\mathrm{m} / \mathrm{s}\) passes through a tank containing 13.5 \(\mathrm{kg}\) of water and emerges with a speed of 534 \(\mathrm{m} / \mathrm{s}\) . What is the maximum temperature increase that the water could have as a result of this event?

Short Answer

Expert verified
The maximum temperature increase of the water is about 0.0087 掳C.

Step by step solution

01

Identify the Known Values

We know the mass of the bullet is 15.0 g (or 0.015 kg), the initial speed of the bullet is 865 m/s, and its final speed after passing through the water is 534 m/s. The mass of the water is 13.5 kg.
02

Calculate the Change in Bullet's Kinetic Energy

The initial kinetic energy of the bullet is \( KE_{initial} = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.015 \times (865)^2 \). The final kinetic energy is \( KE_{final} = \frac{1}{2} \times 0.015 \times (534)^2 \). The change in kinetic energy is \( \Delta KE = KE_{initial} - KE_{final} \).
03

Convert the Energy to Heat Absorbed by Water

The energy lost by the bullet (\( \Delta KE \)) is assumed to be completely absorbed by the water as heat. We set this equal to \( Q = mc\Delta T \), where \( m \) is the mass of the water (13.5 kg), \( c \) is the specific heat capacity of water (4186 J/kg掳C), and \( \Delta T \) is the temperature change.
04

Solve for the Temperature Increase

Rearrange the equation \( Q = mc\Delta T \) to solve for \( \Delta T \): \( \Delta T = \frac{Q}{mc} \). Substitute \( Q = \Delta KE \) from Step 2, and compute the temperature change using the specific heat capacity of water.
05

Calculation

First calculate the initial and final kinetic energies of the bullet and find \( \Delta KE \). Then substitute \( m = 13.5 \), \( c = 4186 \), and \( \Delta KE \) into \( \Delta T = \frac{Q}{mc} \) to find the temperature change.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of this problem, the bullet has kinetic energy because it is moving at high speed. The formula to calculate kinetic energy is \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the object and \( v \) is its velocity.

When the bullet travels through the water, it slows down, losing some of its kinetic energy. This loss occurs due to the resistance from the water, which takes energy from the bullet. This energy must go somewhere; in this case, it is transferred to the water. By calculating the difference in kinetic energy before and after passing through the water, we can determine how much energy the bullet lost. This loss in kinetic energy provides the heat added to the water, which is important for solving the problem.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one substance to another. In this scenario, the energy lost by the bullet as it slows down is transferred to the water as heat. The energy transfer raises the water's temperature, demonstrating the principle that energy cannot be created or destroyed, just transferred or transformed.

When dealing with heat transfer, we use the equation \[ Q = mc \Delta T \] where \( Q \) is the heat transferred, \( m \) is the mass of the substance gaining or losing heat, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Understanding heat transfer is crucial, as it allows us to calculate the resulting temperature increase in the water, using the kinetic energy lost by the bullet.

This concept is essential in thermodynamics as it helps in understanding how energy moves within physical systems.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is required to raise the temperature of a given mass of a substance by one degree Celsius. For water, this value is known and used in calculations related to heat transfer. In our problem, the specific heat capacity of water is 4186 J/kg掳C.

This property differs from one material to another and determines the material's ability to absorb heat compared to others. High specific heat capacity, like that of water, means the substance can absorb a lot of heat without a significant increase in temperature.

Using the formula \[ \Delta T = \frac{Q}{mc} \] where \( Q \) is the heat energy acquired by the water, \( m \) is the mass of the water, and \( c \) is the specific heat capacity, we can find the change in temperature of the water. This calculation reveals how effectively water can absorb the kinetic energy lost by the bullet and convert it into an increase in temperature. This concept highlights how substances respond to the addition of energy, crucial in understanding thermal processes.

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Most popular questions from this chapter

Temperatures in biomedicine. (a) Normal body temperature. The average normal body temperature measured in the mouth is 310 \(\mathrm{K}\) . What would Celsius and Fahrenheit thermometers read for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\) . What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about 7 \(\mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

\bullet A carpenter builds an exterior house wall with a layer of wood 3.0 \(\mathrm{cm}\) thick on the outside and a layer of Styrofoam"" insulation 2.2 \(\mathrm{cm}\) thick on the inside wall surface. The wood has a thermal conductivity of \(0.080 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),\) and the Sty- rofoam TM has a thermal conductivity of 0.010 \(\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) . The interior surface temperature is \(19.0^{\circ} \mathrm{C},\) and the exterior surface temperature is \(-10.0^{\circ} \mathrm{C}\) . (a) What is the temperature at the plane where the wood meets the Styrofoamm? (b) What is the rate of heat flow per square meter through this wall?

". "The Ship of the Desert." Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 -kg camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \((12\) hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\) . (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )\)

\(\bullet\) (a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees and in kelvins? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{Fon}\) January \(23,1916,\) and the next day it plummeted to \(-56.0^{\circ} \mathrm{F} .\) What was the temperature change in Celsius degrees and in kelvins?

Shivering. You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C},\) and it produces approximately 290 \(\mathrm{W}\) of heat power per square meter of body area. \(\mathrm{A} 68 \mathrm{kg}(150 \mathrm{lb}), 1.78 \mathrm{m}\) (5 foot, 10 inch) person has approximately 1.8 \(\mathrm{m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat capacity of the body is about 3500 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) .
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