What is the mass of a sample of liquid water that took 35.4 kJ of heat to apply to it to raise its temperature from 22.5 °C to 55.4 °C? 3.89x10 g 257 g 0.257 g 1.08x10 g a. b. с. d.
What is the mass of a sample of liquid water that took 35.4 kJ of heat to apply to it to raise its temperature from 22.5 °C to 55.4 °C? 3.89x10 g 257 g 0.257 g 1.08x10 g a. b. с. d.
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Question
![### Question:
What is the mass of a sample of liquid water that took 35.4 kJ of heat to apply to it to raise its temperature from 22.5 °C to 55.4 °C?
### Options:
a. \( 3.89 \times 10^3 \) g
b. 257 g
c. 0.257 g
d. \( 1.08 \times 10^3 \) g
---
This question is designed to test your understanding of the relationship between heat energy, mass, and temperature change in a substance, specifically water. The key formula involved is:
\[ q = mc\Delta T \]
Where:
- \( q \) is the heat energy (in this case, 35.4 kJ, which is 35,400 J since 1 kJ = 1000 J)
- \( m \) is the mass of the water (what we need to find)
- \( c \) is the specific heat capacity of water (approximately 4.18 J/g°C)
- \( \Delta T \) is the change in temperature (55.4 °C - 22.5 °C = 32.9 °C)
You can solve for the mass \( m \) using the above formula by rearranging it:
\[ m = \frac{q}{c\Delta T} \]
By plugging in the values, you can determine the correct answer from the given options. This question exemplifies a practical application of heat transfer principles in thermodynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22bc251f-eb72-48c5-9bbb-85b42ccd30ca%2Ff4cf5000-5230-480e-b992-b0d30faec707%2Ffkyjped_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question:
What is the mass of a sample of liquid water that took 35.4 kJ of heat to apply to it to raise its temperature from 22.5 °C to 55.4 °C?
### Options:
a. \( 3.89 \times 10^3 \) g
b. 257 g
c. 0.257 g
d. \( 1.08 \times 10^3 \) g
---
This question is designed to test your understanding of the relationship between heat energy, mass, and temperature change in a substance, specifically water. The key formula involved is:
\[ q = mc\Delta T \]
Where:
- \( q \) is the heat energy (in this case, 35.4 kJ, which is 35,400 J since 1 kJ = 1000 J)
- \( m \) is the mass of the water (what we need to find)
- \( c \) is the specific heat capacity of water (approximately 4.18 J/g°C)
- \( \Delta T \) is the change in temperature (55.4 °C - 22.5 °C = 32.9 °C)
You can solve for the mass \( m \) using the above formula by rearranging it:
\[ m = \frac{q}{c\Delta T} \]
By plugging in the values, you can determine the correct answer from the given options. This question exemplifies a practical application of heat transfer principles in thermodynamics.
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