If 7.3 kJ of energy are required to change the temperature of water from 5.0°C to 70.0°C, what was the volume of water (in mL)? You may assume the density of water is 1.00 g/mL and c= 4.184 J/(g K).
If 7.3 kJ of energy are required to change the temperature of water from 5.0°C to 70.0°C, what was the volume of water (in mL)? You may assume the density of water is 1.00 g/mL and c= 4.184 J/(g K).
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question:**
If 7.3 kJ of energy are required to change the temperature of water from 5.0°C to 70.0°C, what was the volume of water (in mL)? *You may assume the density of water is 1.00 g/mL and \( c_s = 4.184 \, \text{J/g} \, \text{K} \).*
**Explanation:**
To solve this problem, we use the formula for heat transfer:
\[
q = m \cdot c_s \cdot \Delta T
\]
where:
- \( q \) is the heat energy (in joules),
- \( m \) is the mass of the water (in grams),
- \( c_s \) is the specific heat capacity of water (\( 4.184 \, \text{J/g} \, \text{K} \)),
- \( \Delta T \) is the change in temperature (in °C or K).
First, convert the energy from kilojoules to joules:
\[
7.3 \, \text{kJ} = 7300 \, \text{J}
\]
Next, calculate the change in temperature:
\[
\Delta T = 70.0^\circ \text{C} - 5.0^\circ \text{C} = 65.0 \, \text{K}
\]
Rearrange the heat transfer formula to solve for the mass (\( m \)):
\[
m = \frac{q}{c_s \cdot \Delta T} = \frac{7300 \, \text{J}}{4.184 \, \text{J/g} \, \text{K} \cdot 65.0 \, \text{K}}
\]
Calculate \( m \):
\[
m \approx \frac{7300}{272} \approx 26.84 \, \text{g}
\]
Since the density of water is \( 1.00 \, \text{g/mL} \), the volume of water is equal to its mass:
The volume of water is approximately \( 26.84 \, \text{mL} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dbedc34-c1dd-4e16-ba9c-319b01d1eecb%2Febd86782-a661-41b9-bb37-1d6ffacb320d%2F2bz97dr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
If 7.3 kJ of energy are required to change the temperature of water from 5.0°C to 70.0°C, what was the volume of water (in mL)? *You may assume the density of water is 1.00 g/mL and \( c_s = 4.184 \, \text{J/g} \, \text{K} \).*
**Explanation:**
To solve this problem, we use the formula for heat transfer:
\[
q = m \cdot c_s \cdot \Delta T
\]
where:
- \( q \) is the heat energy (in joules),
- \( m \) is the mass of the water (in grams),
- \( c_s \) is the specific heat capacity of water (\( 4.184 \, \text{J/g} \, \text{K} \)),
- \( \Delta T \) is the change in temperature (in °C or K).
First, convert the energy from kilojoules to joules:
\[
7.3 \, \text{kJ} = 7300 \, \text{J}
\]
Next, calculate the change in temperature:
\[
\Delta T = 70.0^\circ \text{C} - 5.0^\circ \text{C} = 65.0 \, \text{K}
\]
Rearrange the heat transfer formula to solve for the mass (\( m \)):
\[
m = \frac{q}{c_s \cdot \Delta T} = \frac{7300 \, \text{J}}{4.184 \, \text{J/g} \, \text{K} \cdot 65.0 \, \text{K}}
\]
Calculate \( m \):
\[
m \approx \frac{7300}{272} \approx 26.84 \, \text{g}
\]
Since the density of water is \( 1.00 \, \text{g/mL} \), the volume of water is equal to its mass:
The volume of water is approximately \( 26.84 \, \text{mL} \).
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