The specific heat capacity of liquid mercury is 0.14 J/g-°C. How many joules of heat are needed to raise the temperature of 6.00 g of mercury from 36.0°C to 75.0°C? Show and upload your work. O 1.7 x 103 j O 33 J 93 J O 6.0 x 10-4 J 1.8 J
The specific heat capacity of liquid mercury is 0.14 J/g-°C. How many joules of heat are needed to raise the temperature of 6.00 g of mercury from 36.0°C to 75.0°C? Show and upload your work. O 1.7 x 103 j O 33 J 93 J O 6.0 x 10-4 J 1.8 J
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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![### Problem Statement
The specific heat capacity of liquid mercury is \(0.14 \, \text{J/g-}^\circ \text{C}\). How many joules of heat are needed to raise the temperature of \(6.00 \, \text{g}\) of mercury from \(36.0^\circ \text{C}\) to \(75.0^\circ \text{C}\)?
**Show and upload your work.**
### Multiple Choice Options
- \( \mathbf{1.7 \times 10^3 \, \text{J}} \)
- \( \mathbf{33 \, \text{J}} \)
- \( \mathbf{93 \, \text{J}} \)
- \( \mathbf{6.0 \times 10^{-4} \, \text{J}} \)
- \( \mathbf{1.8 \, \text{J}} \)
---
### Detailed Explanation
To solve this problem, use the formula for calculating heat energy:
\[ Q = mc\Delta T \]
Where:
- \(Q\) is the heat energy (in joules),
- \(m\) is the mass of the substance (in grams),
- \(c\) is the specific heat capacity (in J/g-°C),
- \(\Delta T\) is the change in temperature (in °C).
1. **Calculate \(\Delta T\) (change in temperature)**:
\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 75.0^\circ \text{C} - 36.0^\circ \text{C} = 39.0^\circ \text{C}
\]
2. **Substitute the values into the formula**:
\[
Q = mc\Delta T = (6.00 \, \text{g}) \times (0.14 \, \text{J/g-}^\circ \text{C}) \times (39.0^\circ \text{C})
\]
3. **Perform the calculation**:
\[
Q = 6.00 \times 0.14 \times 39.0 = 32.76 \, \text{J}
\]
Therefore, the correct answer is approximately \( 33 \, \text{J} \).
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3def50f-0bee-45b6-a067-ec9125eb6917%2Fd2dc47db-099e-449c-b8b9-75362a2c3030%2Foup41p8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
The specific heat capacity of liquid mercury is \(0.14 \, \text{J/g-}^\circ \text{C}\). How many joules of heat are needed to raise the temperature of \(6.00 \, \text{g}\) of mercury from \(36.0^\circ \text{C}\) to \(75.0^\circ \text{C}\)?
**Show and upload your work.**
### Multiple Choice Options
- \( \mathbf{1.7 \times 10^3 \, \text{J}} \)
- \( \mathbf{33 \, \text{J}} \)
- \( \mathbf{93 \, \text{J}} \)
- \( \mathbf{6.0 \times 10^{-4} \, \text{J}} \)
- \( \mathbf{1.8 \, \text{J}} \)
---
### Detailed Explanation
To solve this problem, use the formula for calculating heat energy:
\[ Q = mc\Delta T \]
Where:
- \(Q\) is the heat energy (in joules),
- \(m\) is the mass of the substance (in grams),
- \(c\) is the specific heat capacity (in J/g-°C),
- \(\Delta T\) is the change in temperature (in °C).
1. **Calculate \(\Delta T\) (change in temperature)**:
\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 75.0^\circ \text{C} - 36.0^\circ \text{C} = 39.0^\circ \text{C}
\]
2. **Substitute the values into the formula**:
\[
Q = mc\Delta T = (6.00 \, \text{g}) \times (0.14 \, \text{J/g-}^\circ \text{C}) \times (39.0^\circ \text{C})
\]
3. **Perform the calculation**:
\[
Q = 6.00 \times 0.14 \times 39.0 = 32.76 \, \text{J}
\]
Therefore, the correct answer is approximately \( 33 \, \text{J} \).
-
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