12 ounces of water are heated during the preparation of a cup of coffee. 1150 J of heat are added to the water, which is initially at 20 °C. What is the final temperature of the coffee?
12 ounces of water are heated during the preparation of a cup of coffee. 1150 J of heat are added to the water, which is initially at 20 °C. What is the final temperature of the coffee?
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...
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![**Title: Calculating the Final Temperature of Coffee**
**Introduction:**
When preparing a cup of coffee, understanding the heat transfer involved can help determine the final temperature. In this scenario, we examine how much the temperature of water increases when a specific amount of heat is added.
**Problem Statement:**
12 ounces of water are heated during the preparation of a cup of coffee. Initially, the water is at 20 °C. During the process, 1150 Joules (J) of heat are added to the water. Your task is to calculate the final temperature of the coffee.
**Formula and Approach:**
To find the final temperature of the coffee, use the formula:
\[
q = mc\Delta T
\]
Where:
- \( q \) is the heat added (1150 J)
- \( m \) is the mass of the water
- \( c \) is the specific heat capacity of water (4.18 J/g°C)
- \( \Delta T \) is the change in temperature
**Conversion and Calculation:**
1. Convert ounces to grams:
- 1 ounce = 28.3495 grams
- Therefore, 12 ounces = 12 × 28.3495 = 340.194 grams
2. Plug the values into the formula to find \( \Delta T \):
\[
1150 = 340.194 \times 4.18 \times \Delta T
\]
\[
\Delta T = \frac{1150}{340.194 \times 4.18}
\]
\[
\Delta T \approx 0.811°C
\]
3. Calculate the final temperature:
\[
\text{Final Temperature} = \text{Initial Temperature} + \Delta T
\]
\[
\text{Final Temperature} = 20°C + 0.811°C = 20.811°C
\]
**Conclusion:**
By adding 1150 J of heat to 12 ounces of water initially at 20 °C, the final temperature of the coffee is approximately 20.811 °C. Understanding these principles of heat transfer is essential in practical applications like coffee preparation and in more complex scientific problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f35a07c-a5eb-4d79-bb69-5908bd1ff29b%2Fd047c135-52ba-42aa-b590-4cccaba1d77a%2Flb2tboo_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Final Temperature of Coffee**
**Introduction:**
When preparing a cup of coffee, understanding the heat transfer involved can help determine the final temperature. In this scenario, we examine how much the temperature of water increases when a specific amount of heat is added.
**Problem Statement:**
12 ounces of water are heated during the preparation of a cup of coffee. Initially, the water is at 20 °C. During the process, 1150 Joules (J) of heat are added to the water. Your task is to calculate the final temperature of the coffee.
**Formula and Approach:**
To find the final temperature of the coffee, use the formula:
\[
q = mc\Delta T
\]
Where:
- \( q \) is the heat added (1150 J)
- \( m \) is the mass of the water
- \( c \) is the specific heat capacity of water (4.18 J/g°C)
- \( \Delta T \) is the change in temperature
**Conversion and Calculation:**
1. Convert ounces to grams:
- 1 ounce = 28.3495 grams
- Therefore, 12 ounces = 12 × 28.3495 = 340.194 grams
2. Plug the values into the formula to find \( \Delta T \):
\[
1150 = 340.194 \times 4.18 \times \Delta T
\]
\[
\Delta T = \frac{1150}{340.194 \times 4.18}
\]
\[
\Delta T \approx 0.811°C
\]
3. Calculate the final temperature:
\[
\text{Final Temperature} = \text{Initial Temperature} + \Delta T
\]
\[
\text{Final Temperature} = 20°C + 0.811°C = 20.811°C
\]
**Conclusion:**
By adding 1150 J of heat to 12 ounces of water initially at 20 °C, the final temperature of the coffee is approximately 20.811 °C. Understanding these principles of heat transfer is essential in practical applications like coffee preparation and in more complex scientific problems.
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