/*! 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 96 A \(7.26 \mathrm{kg}\) shot (as ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 96

A \(7.26 \mathrm{kg}\) shot (as used in the sporting event, the shot put) is dropped from the top of a building \(168 \mathrm{m}\) high. What is the maximum temperature increase that could occur in the shot? Assume a specific heat of \(0.47 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\) for the shot. Why would the actual measured temperature increase likely be less than the calculated value?

Short Answer

Expert verified
The maximum temperature increase that could occur in the shot can be calculated by finding the gravitational potential energy at the top of the building and then converting that energy into a temperature increase using the shot's specific heat. The actual temperature increase is likely less than this calculated maximum due to energy losses during the fall and upon impact.

Step by step solution

01

Calculate the gravitational potential energy.

First, calculate the gravitational potential energy using the formula \(PE = m.g.h\). In this case, \(m\) is the mass of the shot, which is equal to \(7.26 \mathrm{kg}\) and \(h\) is the height from which the shot is dropped, which is \(168 \mathrm{m}\). \(g\) refers to the acceleration due to gravity, and we typically use a standard value of \(9.8 \mathrm{m/s}^2\) for this on Earth. The potential energy equation thus becomes \(PE = 7.26 \mathrm{kg} * 9.8 \mathrm{m/s}^2 * 168 \mathrm{m}\).
02

Convert this energy into a temperature increase.

Next, convert the gravitational potential energy into thermal energy using the specific heat capacity formula \(\Delta \theta = \frac {Q} {mc}\), where \(Q\) is the heat absorbed (or potential energy in this case), \(m\) is the mass and \(c\) is the specific heat. Substitute the calculated potential energy into the formula alongside with the known values of \(m\) and \(c\) to find the maximum increase in temperature. Here, \(m = 7.26 \mathrm{kg}\) or \(7260 \mathrm{g}\), \(c = 0.47 \mathrm{Jg}^{-1}\) \(^{\circ}\mathrm{C}^{-1}\), and \(Q\) is the potential energy calculated in previous step.
03

Address why actual temperature would be less.

In practice, the actual temperature increase would likely be less than this calculated value, because not all of the gravitational potential energy will be converted into thermal energy. Some energy may be lost due to air resistance while the shot falls, and more energy will be lost when it hits the ground, as sound, vibration, and potentially even light. These energy losses would reduce the amount of energy that could be converted into heat.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a type of energy stored due to an object's position in a gravitational field, usually relative to the Earth. It's like an energy reserve that objects gain when lifted. The higher you lift something, the more gravitational potential energy it accrues. You can calculate this energy using the formula:
  • \[ PE = mgh \]
where:
  • \(m\) is the mass of the object (in kilograms)
  • \(g\) is the acceleration due to gravity, approximately \( 9.8 \mathrm{m/s}^2 \) on Earth
  • \(h\) is the height above the ground (in meters)
For example, a shot put that weighs \( 7.26 \mathrm{kg} \) and is lifted to a height of \( 168 \mathrm{m} \) will have gravitational potential energy. This stored energy can potentially convert to other forms, such as thermal energy, when the shot put is dropped.
Specific Heat Capacity
Specific heat capacity is a critical concept in understanding how substances absorb heat. It refers to the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. Specific heat capacity can be thought of as a measure of a material's ability to resist temperature change when absorbing heat. For the shot put example:
  • The specific heat capacity of the material of the shot is \(0.47 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).
  • This means it takes \(0.47\) Joules to raise \(1\) gram of the shot material by \(1\) degree Celsius.
Knowing the specific heat capacity is essential to converting potential energy to thermal energy, as it dictates how much temperature change will result from a given amount of energy input. If a substance has a low specific heat capacity, it will change temperature more significantly given the same energy input.
Energy Conservation in Physics
The principle of energy conservation states that energy in a closed system remains constant鈥攊t can neither be created nor destroyed, only transformed from one form to another. This is foundational in physics problems, such as converting potential energy to thermal energy. When the shot put falls, its gravitational potential energy transforms as follows:
  • Some energy is converted to thermal energy, increasing the shot's temperature.
  • Energy is also lost to air resistance, producing sound and potentially causing vibrations.
  • The conservation principle explains why not all potential energy becomes thermal energy; instead, it's partitioned into these various forms.
Thus, the calculated temperature increase reflects an ideal condition where all potential energy becomes heat, while actual conditions lead to energy dispersion through diverse transformations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

A 1.620 g sample of naphthalene, \(C_{10} \mathrm{H}_{8}(\mathrm{s}),\) is completely burned in a bomb calorimeter assembly and a temperature increase of \(8.44^{\circ} \mathrm{C}\) is noted. If the heat of combustion of naphthalene is \(-5156 \mathrm{kJ} / \mathrm{mol} \mathrm{C}_{10} \mathrm{H}_{8}\) what is the heat capacity of the bomb calorimeter?

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.