Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

Short Answer

Expert verified
214.2 grams of steam must condense.

Step by step solution

01

Calculate the heat needed to melt the ice

First, we find the heat needed to melt the 0.450 kg of ice at 0.0°C. The latent heat of fusion of ice is 334,000 J/kg. We calculate the heat (\( Q_1 \)) required: \[ Q_1 = m_\text{ice} \times L_f = 0.450 \, \text{kg} \times 334,000 \, \text{J/kg} = 150,300 \, \text{J} \]
02

Calculate the heat needed to raise the temperature of water from the melted ice and existing water

Next, we find the heat required to raise the temperature of the resulting water (0.450 kg from ice and 2.40 kg existing) from 0.0°C to 28.0°C. The specific heat capacity of water is 4,186 J/kg°C. So, the heat (\( Q_2 \)) required is: \[ Q_2 = (m_\text{ice o water} + m_\text{water}) \times c \times \Delta T = (0.450 \text{ kg} + 2.40 \text{ kg}) \times 4,186 \, \text{J/kg°C} \times 28.0°\text{C} = 333,724.8 \, \text{J} \]
03

Calculate the total heat required for the process

Add the heat needed to melt the ice and raise the temperature to 28.0°C. \[ Q_\text{total} = Q_1 + Q_2 = 150,300 \, \text{J} + 333,724.8 \, \text{J} = 484,024.8 \, \text{J} \]
04

Calculate the mass of steam required

To find the mass of steam needed, use the latent heat of vaporization of water, which is 2,260,000 J/kg. Rearrange the equation for heat to solve for the mass of steam (\( m_\text{steam} \)): \[ Q_\text{total} = m_\text{steam} \times L_v \] Thus, \[ m_\text{steam} = \frac{Q_\text{total}}{L_v} = \frac{484,024.8 \, \text{J}}{2,260,000 \, \text{J/kg}} \approx 0.2142 \, \text{kg} \] Convert this to grams: \[ 0.2142 \, \text{kg} = 214.2 \, \text{g} \]

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 Vaia!

Key Concepts

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

Latent Heat
Latent heat is the energy absorbed or released by a substance during a change of state without changing its temperature. For example, when ice melts into water, it absorbs a certain amount of heat, known as the latent heat of fusion. This energy helps break the molecular bonds in the ice without raising the temperature. It is important to recognize that the amount of latent heat varies depending on the material and the phase change it undergoes.

For ice, the latent heat of fusion is 334,000 J/kg. This means that to melt 1 kg of ice at 0°C into 1 kg of water at 0°C, you need 334,000 joules of heat. No temperature change occurs during this phase; the energy goes into breaking the bonds between water molecules instead of increasing the kinetic energy of the molecules.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to raise the temperature of 1 kilogram of a substance by 1°C. It tells us how a given material responds when it receives heat. For instance, water has a specific heat capacity of 4,186 J/kg°C, which is relatively high compared to many other substances. This high value means water can absorb or lose a lot of heat without experiencing a large temperature change.

Understanding specific heat capacity is crucial in solving thermodynamic problems, such as calculating how much energy is needed to raise the temperature of a mass of water, as we see in thermal exercises where we change the temperature of a system. In our scenario, knowing the specific heat capacity enables us to determine how much energy is necessary not just to melt the ice, but also to warm all water to the desired temperature of 28°C.
Phase Change
A phase change is a transition of matter from one state (solid, liquid, gas) to another. During a phase change, a substance absorbs or releases energy in the form of latent heat, but its temperature remains constant. Common examples of phase changes include melting, freezing, vaporization, and condensation.

Each phase change involves energy exchange without altering the temperature of the substance. In our example, ice melting into water is a phase change where the system absorbs energy to break ice's rigid structure. It’s important to understand the concept of phase changes to calculate how much energy is required to complete such transformations, particularly when combined with temperature changes after melting.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one object or substance to another. This can occur through conduction, convection, or radiation. In thermodynamics problems, understanding how heat is transferred is vital for solving the energy balance within a system.

In our exercise, the heat generated by the condensing steam is transferred to the ice and water in the vessel. The steam loses energy when it condenses, releasing its latent heat of vaporization as it changes from gas to liquid. That energy is then absorbed by the ice and water, leading them to melt and increase in temperature. It's critical to account for all energy transfers in the system to accurately determine the amount of heat required to achieve a desired outcome, such as raising the system's temperature to 28°C.

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 picture window has dimensions of 1.40 m \(\times\) 2.50 mand is made of glass 5.20 mm thick. On a winter day, the temperature of the outside surface of the glass is -20.0\(^\circ\)C, while the temperature of the inside surface is a comfortable 19.5\(^\circ\)C. (a) At what rate is heat being lost through the window by conduction? (b) At what rate would heat be lost through the window if you covered it with a 0.750-mm-thick layer of paper (thermal conductivity 0.0500 W/m \(\cdot\) K)?

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world’s highest average annual temperature (Lugh Ferrandi, Somalia).

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 28.4 N. You carefully add \(1.25 \times 10{^4}\) J of heat energy to the sample and find that its temperature rises 18.0 C\(^\circ\). What is the sample's specific heat?

One end of an insulated metal rod is maintained at 100.0\(^\circ\)C, and the other end is maintained at 0.00\(^\circ\)C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm\(^2\). The heat conducted by the rod melts 8.50 g of ice in 10.0 min. Find the thermal conductivity \(k\) of the metal.

Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m\(^2\) to that for a double-pane window with the same area. The glass of a single pane is 4.2 mm thick, and the air space between the two panes of the double-pane window is 7.0 mm thick. The glass has thermal conductivity 0.80 W /m \(\cdot\) K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m\(^2 \cdot\) K/W.

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.

Sign-up for free