Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/gx°C. O 837 J O 83.7 J O 2.10 x 106 J 47.8 J
Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/gx°C. O 837 J O 83.7 J O 2.10 x 106 J 47.8 J
Chemistry
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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
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![### Heat Calculation in Physics
#### Question:
Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/g×°C.
#### Options:
- 837 J
- 83.7 J
- 2.10 x 10^6 J
- 47.8 J
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This question is part of a physics quiz designed to assess understanding of the concept of specific heat and the calculation of heat transfer. The specific heat formula used to solve this is:
\[ q = mc\Delta T \]
Where:
- \( q \) is the heat energy (in Joules)
- \( m \) is the mass (in grams)
- \( c \) is the specific heat capacity (in J/g×°C)
- \( \Delta T \) is the change in temperature (in °C)
Given data:
- Mass of water, \( m \) = 20.0 g
- Initial temperature, \( T_i \) = 25.1°C
- Final temperature, \( T_f \) = 35.1°C
- Specific heat capacity of water, \( c \) = 4.184 J/g×°C
The temperature change, \( \Delta T = T_f - T_i = 35.1°C - 25.1°C = 10.0°C \).
Now substitute these values into the formula:
\[
q = 20.0 \text{ g} \times 4.184 \text{ J/g} \times 10.0 \text{°C}
\]
\[
q = 837 J
\]
Therefore, the quantity of heat required is **837 Joules**.
This example illustrates the practical application of the specific heat concept and demonstrates how to perform necessary calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed53cb99-e12f-457e-892f-a65c6a7750d0%2F943fab7c-eff8-49fd-bca7-de4f179a5b69%2Fkm3oavo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Heat Calculation in Physics
#### Question:
Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/g×°C.
#### Options:
- 837 J
- 83.7 J
- 2.10 x 10^6 J
- 47.8 J
#### Navigation:
- **Previous Page**: [Button for navigating to the previous page]
- **Next Page**: [Button for navigating to the next page]
#### Quiz Control:
- **Submit Quiz**: [Button to submit the quiz]
- Number of questions saved: 3 out of 30 questions
---
This question is part of a physics quiz designed to assess understanding of the concept of specific heat and the calculation of heat transfer. The specific heat formula used to solve this is:
\[ q = mc\Delta T \]
Where:
- \( q \) is the heat energy (in Joules)
- \( m \) is the mass (in grams)
- \( c \) is the specific heat capacity (in J/g×°C)
- \( \Delta T \) is the change in temperature (in °C)
Given data:
- Mass of water, \( m \) = 20.0 g
- Initial temperature, \( T_i \) = 25.1°C
- Final temperature, \( T_f \) = 35.1°C
- Specific heat capacity of water, \( c \) = 4.184 J/g×°C
The temperature change, \( \Delta T = T_f - T_i = 35.1°C - 25.1°C = 10.0°C \).
Now substitute these values into the formula:
\[
q = 20.0 \text{ g} \times 4.184 \text{ J/g} \times 10.0 \text{°C}
\]
\[
q = 837 J
\]
Therefore, the quantity of heat required is **837 Joules**.
This example illustrates the practical application of the specific heat concept and demonstrates how to perform necessary calculations.
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