In the laboratory a student finds that it takes 921 Joules to increase the temperature of 13.8 grams of liquid water from 20.6 to 37.5 degrees Celsius. The specific heat of water calculated from her data is |J/g°C.
In the laboratory a student finds that it takes 921 Joules to increase the temperature of 13.8 grams of liquid water from 20.6 to 37.5 degrees Celsius. The specific heat of water calculated from her data is |J/g°C.
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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![**Calculating the Specific Heat of Water**
In the laboratory, a student finds that it takes 921 Joules to increase the temperature of 13.8 grams of liquid water from 20.6 to 37.5 degrees Celsius.
The specific heat of water calculated from her data is ______ J/g°C.
You have 8 more group attempts remaining.
Buttons:
- **Submit Answer**
- **Retry Entire Group**
---
To calculate the specific heat (\(c\)) of water, you can use the formula:
\[
q = mc\Delta T
\]
where:
- \(q\) is the amount of heat added (921 Joules)
- \(m\) is the mass of the water (13.8 grams)
- \(\Delta T\) is the change in temperature (37.5°C - 20.6°C = 16.9°C)
- \(c\) is the specific heat, which we need to calculate.
Rearranging the formula to solve for \(c\) gives us:
\[
c = \frac{q}{m\Delta T}
\]
By substituting the known values into the equation, we find:
\[
c = \frac{921 \, \text{J}}{13.8 \, \text{g} \times 16.9 \, \text{°C}}
\]
Calculating this will give you the specific heat of water in J/g°C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9bc6dc7-2424-4f69-956a-285f3752d785%2F45c77dcf-9f96-4d84-9119-2a50e5114243%2F05ibct_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating the Specific Heat of Water**
In the laboratory, a student finds that it takes 921 Joules to increase the temperature of 13.8 grams of liquid water from 20.6 to 37.5 degrees Celsius.
The specific heat of water calculated from her data is ______ J/g°C.
You have 8 more group attempts remaining.
Buttons:
- **Submit Answer**
- **Retry Entire Group**
---
To calculate the specific heat (\(c\)) of water, you can use the formula:
\[
q = mc\Delta T
\]
where:
- \(q\) is the amount of heat added (921 Joules)
- \(m\) is the mass of the water (13.8 grams)
- \(\Delta T\) is the change in temperature (37.5°C - 20.6°C = 16.9°C)
- \(c\) is the specific heat, which we need to calculate.
Rearranging the formula to solve for \(c\) gives us:
\[
c = \frac{q}{m\Delta T}
\]
By substituting the known values into the equation, we find:
\[
c = \frac{921 \, \text{J}}{13.8 \, \text{g} \times 16.9 \, \text{°C}}
\]
Calculating this will give you the specific heat of water in J/g°C.
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