/*! 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 3 How much does the internal energ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How much does the internal energy of \(5.0\) g of ice at precisely \(0{ }^{\circ} \mathrm{C}\) increase as it is changed to water at \(0{ }^{\circ} \mathrm{C}\) ? Neglect the change in volume. The heat needed to melt the ice is $$\Delta Q=m L_{f}=(5.0 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})=400 \mathrm{cal}$$ No external work is done by the ice as it melts and so \(\Delta W=0\). Therefore, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=(400 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})=1.7 \mathrm{~kJ} $$

Short Answer

Expert verified
The internal energy increases by approximately 1.7 kJ.

Step by step solution

01

Understand the Problem

We need to find the increase in internal energy (\( \Delta U \)) when \(5.0\) g of ice at \(0^{\circ} \mathrm{C}\) is converted to water at \(0^{\circ} \mathrm{C}\). This involves calculating the heat absorbed by the ice, using the given specific heat of fusion and the known mass.
02

Use the Formula for Heat Absorption

Calculate the heat absorbed by the ice using the formula \(\Delta Q = m L_f\), where \(m\) is the mass of the ice and \(L_f\) is the latent heat of fusion. Here, \(m = 5.0\) g and \(L_f = 80\) cal/g. So, \(\Delta Q = (5.0 \text{ g})(80 \text{ cal/g}) = 400\text{ cal}\).
03

Apply the First Law of Thermodynamics

According to the First Law of Thermodynamics, \(\Delta Q = \Delta U + \Delta W\), where \(\Delta U\) is the change in internal energy and \(\Delta W\) is the work done. Since no external work is done (\(\Delta W = 0\)), we have \(\Delta U = \Delta Q\). Thus, \(\Delta U = 400\) cal.
04

Convert Units to Joules

Finally, convert the internal energy from calories to joules using the conversion factor \(1 \text{ cal} = 4.184 \text{ J}\). Therefore, \(\Delta U = 400 \text{ cal} \times 4.184 \text{ J/cal} = 1673.6 \text{ J}\) which is approximately \(1.7 \text{ kJ}\).

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.

First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed, only transformed from one form to another. This concept is critical in understanding changes in a system's internal energy. The formula we use for this is:
  • \( \Delta Q = \Delta U + \Delta W \)
where:
  • \( \Delta Q \) represents the heat added to the system
  • \( \Delta U \) is the change in internal energy
  • \( \Delta W \) is the work done by the system
In the context of melting ice in this exercise, the ice absorbs heat (\( \Delta Q \)) without performing work (since \( \Delta W = 0 \)). Thus, the First Law simplifies to \( \Delta Q = \Delta U \). The heat absorbed in this situation is used solely to change the state of the ice from solid to liquid, resulting in a change in internal energy.
Latent Heat of Fusion
The latent heat of fusion is the amount of energy required to change a substance from a solid to a liquid without changing its temperature. This concept is crucial for understanding phase changes, such as when ice melts to become water. For ice, the latent heat of fusion is given as 80 cal/g, a specific heat amount needed to overcome the intermolecular forces holding the ice together.

When solving problems related to heat and phase changes, we use the formula:
  • \( \Delta Q = m L_f \)
where:
  • \( m \) is the mass of the substance
  • \( L_f \) is the latent heat of fusion
In our exercise, we calculated the heat needed to convert 5.0 g of ice to water at the same temperature by multiplying the mass by the latent heat of fusion, resulting in 400 cal. This calculation does not account for any temperature change since the process occurs at a constant \( 0^{\circ} \text{C} \).
Internal Energy Change
The concept of internal energy change is central to thermodynamics. Internal energy refers to the total energy contained within a system, stemming from molecular motion and interactions. When energy in the form of heat is added to the system, as in our exercise where ice turns into water, the internal energy changes.

To find the internal energy change (\( \Delta U \)), we utilize the relationship with heat and work from the First Law of Thermodynamics. Since no work is done by the system in this scenario, the entire amount of heat absorbed (\( \Delta Q = 400 \text{ cal} \)) translates directly into increased internal energy.

We further converted this energy from calories to joules using the conversion factor:
  • \( 1 \text{ cal} = 4.184 \text{ J} \)
Thus, the internal energy increase was calculated to be approximately 1.7 kJ. Understanding this conversion is important for expressing energy in more universally accepted units.

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

By how much does the internal energy of 50 g of oil \((c=0.32 \mathrm{cal} / \mathrm{g}\) \({ }^{\circ} \mathrm{C}\) ) change as the oil is cooled from \(100^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).

A spring \((k=500 \mathrm{~N} / \mathrm{m})\) supports a 400 -g mass, which is immersed in 900 g of water. The specific heat of the mass is \(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The spring is now stretched \(15 \mathrm{~cm}\), and after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped? The energy stored in the spring is dissipated by the effects of friction and goes to heat the water and mass. The energy stored in the stretched spring was $$\mathrm{PE}_{e}=\frac{1}{2} k x^{2}=\frac{1}{2}(500 \mathrm{~N} / \mathrm{m})(0.15 \mathrm{~m})^{2}=5.625 \mathrm{~J}$$ This energy appears as thermal energy that flows into the water and the mass. Using \(\Delta Q=\mathrm{cm} \Delta T\), $$5.625 \mathrm{~J}=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.900 \mathrm{~kg}) \Delta T+(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.40 \mathrm{~kg}) \Delta 7$$ which leads to $$ \Delta T=\frac{5.625 \mathrm{~J}}{3950 \mathrm{~J} / \mathrm{K}}=0.0014 \mathrm{~K} $$

Five moles of neon gas at \(2.00\) atm and \(27.0^{\circ} \mathrm{C}\) is adiabatically compressed to one-third its initial volume. Find the final pressure, final temperature, and external work done on the gas. For neon, \(\gamma\) \(=1.67, c_{v}=0.148 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and \(M=20.18 \mathrm{~kg} / \mathrm{kmol} .\)

Compute the maximum possible efficiency of a heat engine operating between the temperature limits of \(100^{\circ} \mathrm{C}\) and \(400{ }^{\circ} \mathrm{C}\). Remember that our thermodynamic equations are expressed in terms of absolute temperature. The most efficient engine is the Carnot engine, for which $$ \text { Efficiency }=1-\frac{T_{\mathrm{L}}}{T_{\mathrm{H}}}=1-\frac{373 \mathrm{~K}}{673 \mathrm{~K}}=0.446=44.6 \% $$

A 70 -g metal block moving at \(200 \mathrm{~cm} / \mathrm{s}\) slides across a tabletop a distance of \(83 \mathrm{~cm}\) before it comes to rest. Assuming 75 percent of the thermal energy developed by friction goes into the block, how much does the temperature of the block rise? For the metal, \(c=\) \(0.106 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)

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.