3. Calculate the amount of energy (in calories) needed to heat the hot water. Use the heat equation and specific heat of liquid water from Chapter 3. Show your work. ulov ert eleluols Jmlg ser0e.0 el 0 0.00 ts 1elew to ier phow Tuoy wode boleer

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
Question
### Experiment Data: Heating Cold and Hot Water

#### Heating Cold Water:

1. **Mass of Beaker and Cold Water**: 199.74 g
2. **Mass of Beaker**: 107.36 g
3. **Mass of Cold Water**: 92.38 g
4. **Final Temperature of Cold Water**: 90°C
5. **Initial Temperature of Cold Water**: 21°C
6. **Change in Temperature**: 69°C
7. **Elapsed Time**: 324 seconds

#### Heating Hot Water:

1. **Mass of Beaker and Hot Water**: 199.75 g
2. **Mass of Beaker**: 107.36 g
3. **Mass of Hot Water**: 92.39 g
4. **Final Temperature of Hot Water**: 90°C
5. **Initial Temperature of Hot Water**: 45°C
6. **Change in Temperature**: 45°C
7. **Elapsed Time**: 206 seconds

### Summary:
This data represents an experiment comparing the heating of cold and hot water. It details the masses involved, temperature changes, and the time taken to reach final temperatures. The cold water experienced a greater change in temperature over a longer period, while the hot water reached the same final temperature more quickly.
Transcribed Image Text:### Experiment Data: Heating Cold and Hot Water #### Heating Cold Water: 1. **Mass of Beaker and Cold Water**: 199.74 g 2. **Mass of Beaker**: 107.36 g 3. **Mass of Cold Water**: 92.38 g 4. **Final Temperature of Cold Water**: 90°C 5. **Initial Temperature of Cold Water**: 21°C 6. **Change in Temperature**: 69°C 7. **Elapsed Time**: 324 seconds #### Heating Hot Water: 1. **Mass of Beaker and Hot Water**: 199.75 g 2. **Mass of Beaker**: 107.36 g 3. **Mass of Hot Water**: 92.39 g 4. **Final Temperature of Hot Water**: 90°C 5. **Initial Temperature of Hot Water**: 45°C 6. **Change in Temperature**: 45°C 7. **Elapsed Time**: 206 seconds ### Summary: This data represents an experiment comparing the heating of cold and hot water. It details the masses involved, temperature changes, and the time taken to reach final temperatures. The cold water experienced a greater change in temperature over a longer period, while the hot water reached the same final temperature more quickly.
**Question:**

3. Calculate the amount of energy (in calories) needed to heat the hot water. Use the heat equation and specific heat of liquid water from Chapter 3. Show your work.

**Solution Explanation:**

The task is to calculate the energy required to heat water using the specific heat capacity. The equation used is typically:

\[ Q = m \cdot c \cdot \Delta T \]

- **Q** represents the heat energy (in calories).
- **m** is the mass of the water (in grams).
- **c** is the specific heat capacity of water (1 cal/g°C for liquid water).
- **ΔT** is the change in temperature (final temperature - initial temperature, in °C).

**Provided Calculation:**

- The equation written appears to be: 

\[ Q = 45 \, \text{g} \times 1 \, \text{cal/g°C} \times 10 \,^\circ\text{C} \]

The answer to this calculation, indicated in the image, is:

- **Q = 450 calories**

This represents the total energy needed to heat the given amount of water by the specified temperature change.
Transcribed Image Text:**Question:** 3. Calculate the amount of energy (in calories) needed to heat the hot water. Use the heat equation and specific heat of liquid water from Chapter 3. Show your work. **Solution Explanation:** The task is to calculate the energy required to heat water using the specific heat capacity. The equation used is typically: \[ Q = m \cdot c \cdot \Delta T \] - **Q** represents the heat energy (in calories). - **m** is the mass of the water (in grams). - **c** is the specific heat capacity of water (1 cal/g°C for liquid water). - **ΔT** is the change in temperature (final temperature - initial temperature, in °C). **Provided Calculation:** - The equation written appears to be: \[ Q = 45 \, \text{g} \times 1 \, \text{cal/g°C} \times 10 \,^\circ\text{C} \] The answer to this calculation, indicated in the image, is: - **Q = 450 calories** This represents the total energy needed to heat the given amount of water by the specified temperature change.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY