Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 48

An ice cube at \(0.0^{\circ} \mathrm{C}\) is slowly melting. What is the change in the ice cube's entropy for each \(1.00 \mathrm{g}\) of ice that melts?

Short Answer

Expert verified
Answer: The change in entropy for every 1.00 g of ice that melts from 0.0°C is approximately 1.22 J/K.

Step by step solution

01

Determine the heat of fusion of ice

The heat of fusion is the amount of heat required to change a substance from the solid phase to the liquid phase without changing its temperature. For ice, the heat of fusion is equal to \(333.55 \mathrm{J/g}\).
02

Calculate the heat needed for melting 1 gram of ice

For every \(1.00 \mathrm{g}\) of ice that melts, we can determine the total heat needed for the process by multiplying the heat of fusion by the mass of ice. $$Q = m \times L = (1.00 \thinspace \mathrm{g}) \times (333.55 \thinspace \mathrm{J/g}) = 333.55 \thinspace \mathrm{J}$$
03

Calculate the change in entropy

To calculate the change in entropy, we will use the formula: $$\Delta S = \frac{Q}{T}$$ where \(\Delta S\) is the change in entropy, \(Q\) is the heat added to the system, and \(T\) is the temperature. Remember to use temperature in Kelvin. First, convert the temperature to Kelvin: $$T_{\text{K}} = T_{\text{C}} + 273.15 = 0.0^{\circ} \mathrm{C} + 273.15 \thinspace\mathrm{K} = 273.15 \thinspace\mathrm{K}$$ Now, calculate the change in entropy: $$\Delta S = \frac{333.55 \thinspace \mathrm{J}}{273.15 \thinspace \mathrm{K}} \approx 1.22 \thinspace \mathrm{J/K}$$ So, for every \(1.00 \mathrm{g}\) of ice that melts, the change in the ice cube's entropy is approximately \(1.22 \thinspace \mathrm{J/K}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

An ideal gas is heated at a constant pressure of $2.0 \times 10^{5} \mathrm{Pa}\( from a temperature of \)-73^{\circ} \mathrm{C}$ to a temperature of \(+27^{\circ} \mathrm{C}\). The initial volume of the gas is $0.10 \mathrm{m}^{3} .\( The heat energy supplied to the gas in this process is \)25 \mathrm{kJ} .$ What is the increase in internal energy of the gas?
A balloon contains \(200.0 \mathrm{L}\) of nitrogen gas at $20.0^{\circ} \mathrm{C}$ and at atmospheric pressure. How much energy must be added to raise the temperature of the nitrogen to \(40.0^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure?
Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$
A heat engine takes in \(125 \mathrm{kJ}\) of heat from a reservoir at $815 \mathrm{K}\( and exhausts \)82 \mathrm{kJ}\( to a reservoir at \)293 \mathrm{K}$ (a) What is the efficiency of the engine? (b) What is the efficiency of an ideal engine operating between the same two reservoirs?
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Radiation

Read Explanation

Wave Optics

Read Explanation

Nuclear Physics

Read Explanation

Oscillations

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.

Sign-up for free