/*! 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 127 An open cooking pot containing \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 127

An open cooking pot containing \(0.5\) liter of water at \(20^{\circ} \mathrm{C}, 1\) bar sits on a stove burner. Once the burner is turned on, the water is gradually heated at a rate of \(0.85 \mathrm{~kW}\) while pressure remains constant. After a period of time, the water starts boiling and continues to do so until all of the water has evaporated. Determine (a) the time required for the onset of evaporation, in \(\mathrm{s}\). (b) the time required for all of the water to evaporate, in s, once evaporation starts.

Short Answer

Expert verified
196.7 s for onset of evaporation. 1329.4 seconds for complete evaporation.

Step by step solution

01

- Find the specific heat capacity of water

First, determine the specific heat capacity of water, which is approximately \(4.18 \ \text{J/g}\text{K}\).
02

- Convert mass of water to grams

Next, convert the volume of water into mass. Given that the density of water is approximately 1 g/mL, 0.5 liters (500 mL) will have a mass of \(500 \ \text{g}\).
03

- Determine the heat required for onset of evaporation

Use the formula \(Q = mc\triangle T\) to determine the amount of heat energy (\text{Q}) required to raise the temperature of the water from 20掳C to 100掳C (the boiling point).\(m = 500 \text{g}\) is the mass of the water\(c = 4.18 \text{J/g\text{K}}\) is the specific heat capacity of water\(\triangle T = 100^\text{circ}C - 20^\text{circ}C = 80^\text{circ}C\) is the change in temperature\Thus,\(Q = 500 \text{g}\times 4.18 \text{J/g}\text{K}) \times 80^\text{circ}C) = 167,200\text{J}\)
04

- Calculate the time for onset of evaporation

To find the time required to reach the onset of evaporation, use the formula\(t = \frac{Q}{P}\), where \(P = 0.85 \text{kW} = 850\text{J/s}\). Thus,\(t = \frac{167,200\text{J}}{850\text{J/s}} = 196.7\text{s}\).
05

- Determine the heat required for complete evaporation

Next, calculate the heat required to convert all of the water into vapor using the heat of vaporization of water, which is approximately \(2260 \text{J/g}\). Thus,\(Q_{\text{vaporization}} = 500 \text{g} \times 2260 \text{J/g} = 1,130,000 \text{J}\).
06

- Calculate the time for complete evaporation

To find the time required for complete evaporation, use the formula\(t = \frac{Q_{\text{vaporization}}}{P}\). Thus,\(t = \frac{1,130,000\text{J}}{850\text{J/s}} = 1329.4\text{s}\).

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 91影视!

Key Concepts

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

specific heat capacity
Specific heat capacity refers to the amount of heat needed to increase the temperature of a unit mass of a substance by one degree Celsius. For water, this value is found to be approximately 4.18 J/g掳C. Understanding specific heat capacity is essential because it allows us to calculate the amount of heat energy required to heat water from one temperature to another. These calculations are crucial in thermodynamics problems, especially when determining how much energy is needed to reach the boiling point.
heat energy calculation
Heat energy calculation involves determining the amount of energy needed to change the temperature of a substance. This is done using the formula:
\( Q = mc\Delta T \), where:
  • \( Q \) is the heat energy
  • \( m \) is the mass of the substance
  • \( c \) is the specific heat capacity
  • \( \Delta T \) is the change in temperature

In this exercise, you used this formula to find out how much heat energy was needed to raise the temperature of 500 g of water from 20掳C to 100掳C. By plugging in the values, \( Q = 500\ g \times 4.18\ J/g掳C \times 80掳C = 167,200\ J \), you found the heat energy required.
heat of vaporization
Heat of vaporization is the amount of heat needed to convert a unit mass of a liquid into vapor without changing its temperature. For water, this value is around 2260 J/g. Understanding this concept is essential for determining how much energy is needed to completely convert water to steam. In the calculation, you determined this by using the formula:
\( Q_{\text{vaporization}} = mH_{\text{vaporization}} \), where:
  • \( Q_{\text{vaporization}} \) is the heat energy
  • \( m \) is the mass of the substance
  • \( H_{\text{vaporization}} \) is the heat of vaporization

By plugging in the values, you got \( Q_{\text{vaporization}} = 500\ g \times 2260\ J/g = 1,130,000 J \).
mass and volume conversion
Converting mass and volume is a fundamental step in solving thermodynamics problems. Water's density of approximately 1 g/mL means that its mass can be directly converted from volume. In this exercise, the volume of water was given as 0.5 liters, or 500 mL. Consequently, the mass in grams is equal to the volume in milliliters, hence 500 mL = 500 g.

This conversion is crucial as it allows us to deploy thermodynamic formulas where mass must be used.
constant pressure process
In thermodynamics, a constant pressure process is one where the pressure remains unchanged throughout. This concept is significant when dealing with heating and phase changes, as it simplifies the calculations.

In the given problem, the water was heated at a constant pressure of 1 bar. As a result, the calculations for the energy required to change the state of water from liquid to vapor could be done without having to adjust for varying pressure. This makes understanding the constant pressure process vital for accurate problem-solving in thermodynamics.

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

Nitrogen gas is contained in a rigid \(1-\mathrm{m}\) tank, initially at 10 bar, \(300 \mathrm{~K}\). Heat transfer to the contents of the tank occurs until the temperature has increased to \(400 \mathrm{~K}\). During the process, a pressure-relief valve allows nitrogen to escape, maintaining constant pressure in the tank. Neglecting kinetic and potential energy effects, and using the ideal gas model with constant specific heats evaluated at \(350 \mathrm{~K}\), determine the mass of nitrogen that escapes, in \(\mathrm{kg}\), and the amount of energy transfer by heat, in \(\mathrm{kJ}\).

Steam enters a counterflow heat exchanger operating at steady state at \(0.07 \mathrm{MPa}\) with a specific enthalpy of \(2431.6 \mathrm{~kJ} / \mathrm{kg}\) and exits at the same pressure as saturated liquid. The steam mass flow rate is \(1.5 \mathrm{~kg} / \mathrm{min}\). A separate stream of air with a mass flow rate of \(100 \mathrm{~kg} / \mathrm{min}\) enters at \(30^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The ideal gas model with \(c_{p}=\) \(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) can be assumed for air. Kinetic and potential energy effects are negligible. Determine (a) the quality of the entering steam and (b) the rate of heat transfer between the heat exchanger and its surroundings, in \(\mathrm{kW}\).

Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at \(270 \mathrm{~K}\) with a velocity of \(180 \mathrm{~m} / \mathrm{s}\) and exits with a velocity of \(48.4 \mathrm{~m} / \mathrm{s}\). For negligible potential energy effects, determine the exit temperature, in \(\mathrm{K}\).

The procedure to inflate a hot-air balloon requires a fan to move an initial amount of air into the balloon envelope followed by heat transfer from a propane burner to complete the inflation process. After a fan operates for 10 minutes with negligible heat transfer with the surroundings, the air in an initially deflated balloon achieves a temperature of \(80^{\circ} \mathrm{F}\) and a volume of \(49,100 \mathrm{ft}^{3}\). Next the propane burner provides heat transfer as air continues to flow into the balloon without use of the fan until the air in the balloon reaches a volume of \(65,425 \mathrm{ft}^{3}\) and a temperature of \(210^{\circ} \mathrm{F}\). Air at \(77^{\circ} \mathrm{F}\) and \(14.7 \mathrm{lb} / 1 n^{2}\) surrounds the balloon. The net rate of heat transfer is \(7 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Ignoring effects due to kinetic and potential energy, modeling the air as an ideal gas, and assuming the pressure of the air inside the balloon remains the same as that of the surrounding air, determine (a) the power required by the fan, in hp. (b) the time required for full inflation of the balloon, in min.

Air enters a horizontal, constant-diameter heating duct operating at steady state at \(290 \mathrm{~K}, 1\) bar, with a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\), and exits at \(325 \mathrm{~K}, 0.95\) bar. The flow area is \(0.04 \mathrm{~m}^{2}\). Assuming the ideal gas model with \(k=1.4\) for the air, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), (b) the velocity at the inlet and exit, each in \(\mathrm{m} / \mathrm{s}\), and (c) the rate of heat transfer, in \(\mathrm{kW}\).
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Linear Momentum

Read Explanation

Measurements

Read Explanation

Force

Read Explanation

Thermodynamics

Read Explanation

Translational Dynamics

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.