/*! 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 36 A \(150 \mathrm{~g}\) copper bow... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(150 \mathrm{~g}\) copper bowl contains \(220 \mathrm{~g}\) of water, both at \(20.0^{\circ} \mathrm{C}\). A very hot \(300 \mathrm{~g}\) copper cylinder is dropped into the water, causing the water to boil, with \(5.00 \mathrm{~g}\) being converted to steam. The final temperature of the system is \(100^{\circ} \mathrm{C}\). Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder?

Short Answer

Expert verified
(a) 20300 calories, (b) 1108.8 calories, (c) 871.4°C.

Step by step solution

01

Understand the system and initial conditions

We have a system involving a copper bowl, water, and a copper cylinder. The initial temperature of the bowl and the water is \(20.0^{\circ} \mathrm{C}\). The copper cylinder's temperature is unknown, and it is very hot. The water and the bowl eventually reach \(100^{\circ} \mathrm{C}\), with 5 g of water converted to steam.
02

Calculate energy needed to heat water and vaporize some as steam

First, we calculate the energy needed to raise the water from \(20.0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\), using \(Q = m c \Delta T\), where \(m = 220 \mathrm{~g}\), \(c = 1 \text{ calorie/g°C}\), and \(\Delta T = 80^{\circ} \mathrm{C}\). This gives \(Q_1 = 220 \times 1 \times 80 = 17600\text{ calories}\). Additionally, to convert 5 g of water to steam, we use the latent heat of vaporization \(540 \text{ calories/g}\), leading to \(Q_2 = 5 \times 540 = 2700\text{ calories}\). Total energy transferred to water is \(Q_{\text{water}} = Q_1 + Q_2 = 17600 + 2700 = 20300\text{ calories}\).
03

Calculate energy transferred to the copper bowl

Use the formula \(Q = m c \Delta T\) for the copper bowl. With \(m = 150 \mathrm{~g}\), \(c = 0.0924 \text{ calorie/g°C}\) (specific heat of copper), and \(\Delta T = 80^{\circ} \mathrm{C}\), we get \(Q_{\text{bowl}} = 150 \times 0.0924 \times 80 = 1108.8 \text{ calories}\).
04

Analyze energy conservation to find original cylinder temperature

Energy conservation dictates that the energy lost by the copper cylinder equals the total energy gained by the water and bowl. \(Q_{\text{cylinder}} = Q_{\text{water}} + Q_{\text{bowl}} = 20300 + 1108.8 = 21408.8\text{ calories}\). Now, we backtrack the cylinder's initial temperature using \(Q = m c \Delta T\) for copper, with \(m = 300 \mathrm{~g}\), \(c = 0.0924 \text{ calorie/g°C}\), and \(\Delta T = T_{\text{initial}} - 100\). So, solve \(21408.8 = 300 \times 0.0924 \times (T_{\text{initial}} - 100)\) which simplifies to \(T_{\text{initial}} = \frac{21408.8}{27.72} + 100\). Thus, \(T_{\text{initial}} \approx 871.4^{\circ} \mathrm{C}\).

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 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer
In thermodynamics, heat transfer is the process of energy moving from a hotter object to a cooler one. Understanding how heat moves is crucial in solving problems like the one presented. In the exercise, heat was transferred from a hot copper cylinder to a cooler copper bowl and water. The cylinder, being initially very hot, cooled down as it lost heat energy while the bowl and water, starting cooler, absorbed that heat energy, eventually reaching the boiling point of water, which is 100°C.

Heat transfer factors heavily into many real-world applications. Specifically, in the exercise, we can see it described through formulas such as:
  • Heat added to the water, causing temperature changes and phase changes (from liquid to steam).
  • Heat absorbed by the copper bowl, raising its temperature.
It's essential to monitor these exchanges to use energy efficiently and effectively in various processes.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is required to change the temperature of a specific mass of a substance by one degree Celsius. It's a fundamental concept in thermodynamics as it determines how substances respond to heat energy inputs and outputs. In the exercise, we used specific heat capacity to calculate the amount of energy transferred to both the water and the copper bowl.

The specific heat capacity values are crucial:
  • For water, it's usually 1 calorie/g°C, which denotes the high energy requirement to change its temperature, explaining why bodies of water hold temperature well.
  • For copper, it's 0.0924 calorie/g°C, a much lower value, indicating it heats up and cools down quicker compared to water.
These properties dictate how substances respond in thermal environments, influencing design considerations in engineering and environmental management.
Latent Heat of Vaporization
Latent heat of vaporization is the heat required for a phase change from liquid to gas without changing temperature. It's a concept that emerges when considering steam formation in the exercise. To convert water to steam, energy is required in addition to simply raising its temperature.

In our exercise, 5 grams of water turned into steam. To calculate this energy, we employed the latent heat of vaporization for water, which is 540 calories/g. The calculation was straightforward:
  • 5 g of water into steam: demanded an energy input of 2700 calories (5 g × 540 calories/g).
This energy doesn't raise the temperature but changes the phase. Comprehending latent heat helps in understanding phenomena like boiling and condensation that are part of various industrial and natural processes.
Energy Conservation
Energy conservation is a principle stating that energy cannot be created or destroyed, only transferred or transformed. In thermodynamic exercises like the provided one, this principle plays a significant role in discovering unknown variables.
In the given problem, the total energy lost by the hot cylinder should equal the total energy gained by the bowl and water. This was calculated as:
  • Total energy transferred to the water and bowl: 21408.8 calories.
To find the original temperature of the cylinder, this total energy value was used. The exercise requires solving equations to ensure total heat loss equals total heat gain, adhering to energy conservation principles.
Recognizing energy conservation is crucial for problem-solving and evaluating energy flow effectiveness in complex systems, like power plants and engines, impacting how we develop sustainable energy solutions.

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

The average rate at which energy is conducted outward through the ground surface in North America is \(54.0 \mathrm{~mW} / \mathrm{m}^{2}\), and the average thermal conductivity of the near-surface rocks is \(2.50\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). Assuming a surface temperature of \(10.0{ }^{\circ} \mathrm{C}\), find the temperature at a depth of \(35.0 \mathrm{~km}\) (near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

A rectangular plate of glass initially has the dimensions \(0.200 \mathrm{~m}\) by \(0.300 \mathrm{~m}\). The coefficient of linear expansion for the glass is \(9.00 \times 10^{-6} / \mathrm{K}\). What is the change in the plate's area if its temperature is increased by \(20.0 \mathrm{~K}\) ?

The specific heat of a substance varies with temperature according to the function \(c=0.20+0.14 T+0.023 T^{2}\), with \(T\) in \({ }^{\circ} \mathrm{C}\) and \(c\) in cal \(/ \mathrm{g} \cdot \mathrm{K}\). Find the energy required to raise the temperature of \(2.0 \mathrm{~g}\) of this substance from \(5.0^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).

Nonmetric version: (a) How long does a \(2.0 \times 10^{5} \mathrm{Btu} / \mathrm{h}\) water heater take to raise the temperature of 40 gal of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: (b) How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

An insulated Thermos contains \(130 \mathrm{~cm}^{3}\) of hot coffee at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

See all solutions

Recommended explanations on Physics Textbooks

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.