91Ó°ÊÓ

A industrious explorer of the polar regions has devised a contraption for melting ice. It consists of a sealed \(10 \mathrm{~L}\) cylindrical tank with a porous grate separating the top half from the bottom half. The bottom half includes a paddle wheel attached to an axle that passes outside the cylinder, where it is attached by a gearbox and pulley system to a stationary bicycle. Pedaling the bicycle rotates the paddle wheel inside the cylinder. The tank includes \(6.00 \mathrm{~L}\) of water and \(3.00 \mathrm{~kg}\) of ice at \(0.0^{\circ} \mathrm{C}\). The water fills the bottom chamber, where it may be agitated by the paddle wheel, and partially fills the upper chamber, which also includes the ice. The bicycle is pedaled with an average torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) at a rate of 30.0 revolutions per minute. The system is \(70 \%\) efficient. (a) For what length of time must the explorer pedal the bicycle to melt all the ice? (b) How much longer must he pedal to raise the temperature of the water to \(10.5^{\circ} \mathrm{C}\) ?

Step by step solution

01

Calculate the total work done

First, calculate the total work done by the pedaling. The power provided by the bicycle is given by \($ P = \tau \cdot \omega \), where \( \tau = 25.0 \, \mathrm{N \cdot m} \) is the torque and \( \omega \) is the angular speed. The angular speed can be calculated from the given rate of 30.0 revolutions per minute as \( \omega = 2\pi \cdot \frac{30}{60} \, \mathrm{s^{-1}} \). The power is therefore \( P = 25.0 \, \mathrm{N \cdot m} \cdot 2\pi \cdot \frac{30}{60} \, \mathrm{s^{-1}} = 78.5 \, \mathrm{W} \). The total work \($ W \) done over a time \( t \) is then just \( W = P \cdot t \). Given the efficiency \(\eta = 0.7\), the useful work is \(W_{\text{useful}} = \eta \cdot W\).

02

Calculate the energy required to melt the ice

Next, calculate how much energy is needed to melt the ice. The energy \(Q_{\text{melt}}\) required to melt a mass \(m\) of ice is given by \(Q_{\text{melt}} = m \cdot L_{\text{f}}\), where \(L_{\text{f}} = 334 \, \mathrm{kJ/kg}\) is the latent heat of fusion of ice. Substituting \(m = 3.00 \, \mathrm{kg}\) gives \(Q_{\text{melt}} = 3.00 \cdot 334 \, \mathrm{kJ} = 1000.2 \, \mathrm{kJ}\). From this, the time required to melt the ice can be found by solving the equation \(W_{\text{useful}} = Q_{\text{melt}}\) for \(t\).

03

Calculate the energy required to heat the water

Similarly, the energy \(Q_{\text{heat}}\) required to raise the temperature of the water by \(\Delta T = 10.5 \, \mathrm{C}\) is given by \(Q_{\text{heat}} = ms_{\text{water}}\Delta T\), where \(s_{\text{water}} = 4.184 \, \mathrm{kJ/kg \cdot C}\) is the specific heat of water and \(m = 3.00 \, \mathrm{kg} + \frac{6.00}{1.00} \, \mathrm{kg}\) is the total mass of water, including that from the melted ice. This gives \(Q_{\text{heat}} = 37.76 \, \mathrm{kJ}\). The additional time required to heat the water can then be found by solving the equation \(W_{\text{useful}} = Q_{\text{heat}}\) for \(t\).

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.

Latent Heat of Fusion

When we talk about the latent heat of fusion, we're exploring how a substance changes its state from solid to liquid. In the world of thermodynamics, this concept has a very pivotal role. For the ice to transition into water, it requires energy. But this energy doesn't raise the temperature until the state change is complete. It only goes into changing the structure of the ice.
In our example, the explorer is using a mechanism to melt ice. The amount of energy needed is calculated using the formula:\[ Q_{\text{melt}} = m \cdot L_{\text{f}} \]Where:

• \( m \) is the mass of the ice, in kilograms.
• \( L_{\text{f}} \) is the latent heat of fusion, a specific constant for each material. For ice, it’s 334 kJ/kg.

This tells us the energy required to convert ice at its melting point into water fully.

Specific Heat Capacity

After the ice melts, the next task is to raise the temperature of this water to a desired level. This is where the concept of specific heat capacity comes into play. Specific heat capacity talks about how much energy is needed to change the temperature of a unit mass of a substance by one degree Celsius.
For water, the specific heat capacity is 4.184 kJ/kg°C. So to calculate the energy required to raise the water's temperature, we use the formula:\[ Q_{\text{heat}} = m \cdot s_{\text{water}} \cdot \Delta T \]Where:

• \( m \) is the total mass of the water.
• \( s_{\text{water}} \) is the specific heat capacity of water.
• \( \Delta T \) represents the change in temperature.

This equation helps us understand how different substances require different amounts of energy for temperature changes.

Efficiency of Heat Engines

Efficiency is all about how well a system converts input energy into productive work. In thermodynamics, a heat engine's efficiency can vary widely, depending on the setup and materials used.
In the explorer's scenario, this efficiency is given as 70%. This means that not all the energy from pedaling the bicycle goes into melting the ice or heating the water. Some energy is lost to the environment.
To find the useful work output, we apply the efficiency to the total work done:\[ W_{\text{useful}} = \eta \cdot W \]Where:

• \( \eta \) is the efficiency of the system (70% or 0.7).
• \( W \) is the total work done.

Understanding efficiency helps us design better systems and reminds us of the importance of minimizing energy lost to waste.

Energy Transfer Calculations

Energy transfer is a central discussion in thermodynamics, focusing on how energy moves from one place to another. The explorer's device is a brilliant example of converting mechanical energy into heat to change the ice's state and temperature.
The whole process involves calculating various forms of energy and how they translate from mechanical work to heat energy, with efficiency playing a major role.
From calculating the power output of pedaling:\[ P = \tau \cdot \omega \]Where:

• \( \tau \) is the torque applied.
• \( \omega \) is the angular velocity.

To figuring out useful work and energy required for phase and temperature changes, we see energy transfer concepts interwoven in every step. Understanding these calculations allows us to predict outcomes and fine-tune devices for optimal performance.