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Chapter 17: Problem 37

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 approximately 0.061 掳C.

Step by step solution

01

Determine Initial and Final Kinetic Energy of Bullet

The initial kinetic energy (KE) of the bullet is given by:\[ KE_{initial} = \frac{1}{2} m v_{i}^2 \]Substitute the given values:\[ m = 0.015 \text{ kg}, \quad v_i = 865 \text{ m/s} \]\[ KE_{initial} = \frac{1}{2} \times 0.015 \text{ kg} \times (865 \text{ m/s})^2 \approx 5612.8125 \text{ J} \]The final kinetic energy (KE) after passing through water is:\[ KE_{final} = \frac{1}{2} m v_{f}^2 \]Substitute the given values:\[ v_f = 534 \text{ m/s} \]\[ KE_{final} = \frac{1}{2} \times 0.015 \text{ kg} \times (534 \text{ m/s})^2 \approx 2138.07 \text{ J} \]
02

Calculate the Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is the difference between the initial and final kinetic energies:\[ \Delta KE = KE_{initial} - KE_{final} \]Substituting the found values:\[ \Delta KE = 5612.8125 \text{ J} - 2138.07 \text{ J} \approx 3474.7425 \text{ J} \]
03

Relate the Change in Kinetic Energy to Heat Energy

Assuming no other losses, the change in kinetic energy of the bullet is transferred as heat energy \( Q \) to the water:\[ Q = \Delta KE \approx 3474.7425 \text{ J} \]
04

Calculate the Temperature Increase of Water

The increase in temperature \( \Delta T \) of the water is given by:\[ Q = mc\Delta T \]Where \( m = 13.5 \text{ kg} \) is the mass of the water, \( c = 4184 \text{ J/kg}^\circ\text{C} \) is the specific heat capacity of water, and \( Q = 3474.7425 \text{ J} \).Rearranging for \( \Delta T \):\[ \Delta T = \frac{Q}{mc} = \frac{3474.7425 \text{ J}}{13.5 \text{ kg} \times 4184 \text{ J/kg}^\circ\text{C}} \approx 0.061 \text{ } ^\circ\text{C} \]
05

Conclusion

Considering energy conservation and heat transfer principles, the maximum possible temperature increase for the water is approximately 0.061 掳C.

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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 that explains how thermal energy moves from one place or substance to another. In the context of the bullet passing through water, the energy that is transferred is primarily in the form of heat. When the bullet enters the water, it loses some of its kinetic energy. This lost energy doesn鈥檛 just disappear; it is transferred as heat to the water molecules.

In practical terms, heat transfer can occur in several ways, including conduction, convection, and radiation. For this particular scenario, conduction plays a significant role since the bullet is in direct contact with the water. The heat from the bullet is conducted into the water quietly, which slightly raises the water's temperature.

Heat transfer is crucial in our daily lives, from cooking food to weather systems. Understanding this helps us appreciate how energy moves and changes form.
Specific Heat Capacity
Specific heat capacity is a property that indicates how much heat energy is needed to change the temperature of a substance. Specifically, it refers to the amount of energy required to increase the temperature of one kilogram of a substance by one degree Celsius.

In the given problem, the specific heat capacity of water is used. Water has a relatively high specific heat capacity of 4184 J/kg掳C, meaning it can absorb a lot of heat without a significant change in temperature. This property is why oceans stabilize climates and help regulate temperature fluctuations on Earth.

In our calculation, the specific heat capacity helped us determine how much the water's temperature rose when it absorbed the kinetic energy lost by the bullet. This real-world application shows how specific heat capacity is essential for tasks involving heating and cooling.
Energy Conservation
Energy conservation is a crucial principle in physics. It states that in an isolated system, energy cannot be created or destroyed, only transformed from one form to another. This principle helps us analyze scenarios, such as the bullet passing through water, with accuracy and precision.

Initially, the bullet has a high kinetic energy. As it enters and exits the water tank, some of this kinetic energy is converted into heat energy, which causes a slight rise in the water's temperature. The concept of energy conservation allows us to calculate this change by understanding that the energy lost by the bullet equals the heat absorbed by the water (assuming no other losses).

By conserving energy throughout the process, scientists and engineers can develop more accurate models for a wide range of phenomena, helping them design efficient systems and processes in various fields like thermodynamics, engineering, and environmental science.

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Most popular questions from this chapter

(a) A wire that is 1.50 \(\mathrm{m}\) long at \(20.0^{\circ} \mathrm{C}\) is found to increase in length by 1.90 \(\mathrm{cm}\) when warmed to \(420.0^{\circ} \mathrm{C}\) . Compute its average coefficient of linear expansion for this temperature range. (b) The wire is stretched just taut (zero tension) at \(420.0^{\circ} \mathrm{C}\). Find the stress in the wire if it is cooled to \(20.0^{\circ} \mathrm{C}\) without being allowed to contract. Young's modulus for the wire is \(2.0 \times 10^{11} \mathrm{Pa}\)

(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? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{F} .\) What was the temperature change in Celsius degrees?

A gas thermometer registers an absolute pressure corresponding to 325 \(\mathrm{mm}\) of mercury when in contact with water at the triple point. What pressure does it read when in contact with water at the normal boiling point?

A crate of fruit with mass 35.0 \(\mathrm{kg}\) and specific heat 3650 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) slides down a ramp inclined at \(36.9^{\circ}\) below the horizontal. The ramp is 8.00 \(\mathrm{m}\) long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

Heat Loss During Breathing. In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat of air is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that 1.0 \(\mathrm{L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{kg}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?
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