A chemist carefully measures the amount of heat needed to raise the temperature of a 0.53 kg sample of a pure substance from 10.7 °C to 16.6 °C. The experiment shows that 0.73 kJ of heat are needed. What can the chemist report for the specific heat capacity of the substance? Be sure your answer has the correct number of significant digits.
A chemist carefully measures the amount of heat needed to raise the temperature of a 0.53 kg sample of a pure substance from 10.7 °C to 16.6 °C. The experiment shows that 0.73 kJ of heat are needed. What can the chemist report for the specific heat capacity of the substance? Be sure your answer has the correct number of significant digits.
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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![**Title: Determining the Specific Heat Capacity of a Substance**
**Introduction:**
Understanding the specific heat capacity of substances is crucial in the field of chemistry. It allows us to predict how a substance will react to changes in temperature when energy is added or removed. This experiment demonstrates how to measure and calculate the specific heat capacity of a given substance.
**Problem Statement:**
A chemist carefully measures the amount of heat needed to raise the temperature of a 0.53 kg sample of a pure substance from 10.7°C to 16.6°C. The experiment shows that 0.73 kJ of heat (Q) are needed. What can the chemist report for the specific heat capacity (c) of the substance? Make sure your answer has the correct number of significant digits.
**Given Data:**
- Mass of the substance (m): 0.53 kg
- Initial temperature (T1): 10.7°C
- Final temperature (T2): 16.6°C
- Heat added (Q): 0.73 kJ
**Objective:**
Calculate the specific heat capacity (c) of the substance using the given data.
**Formula:**
To determine the specific heat capacity, use the formula:
\[ c = \frac{Q}{m \cdot \Delta T} \]
where:
- \( Q \) is the amount of heat added (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( \Delta T \) is the change in temperature (in Kelvin/Celsius since the change is the same in both).
**Steps to Solve:**
1. Convert the heat added from kJ to J:
\[ 0.73 \, \text{kJ} = 0.73 \times 1000 \, \text{J} = 730 \, \text{J} \]
2. Calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T2 - T1 = 16.6°C - 10.7°C = 5.9°C \]
3. Substitute the values into the formula:
\[ c = \frac{730 \, \text{J}}{0.53 \, \text{kg} \times 5.9 \, \text{°C}} \]
4. Compute the result:
\[ c = \frac{730}{0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd290e29f-bf0a-4e13-893a-d0d06da948d8%2F2b8efedb-5414-442d-918c-5f03bf9ea8b1%2Fgyeqw5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Determining the Specific Heat Capacity of a Substance**
**Introduction:**
Understanding the specific heat capacity of substances is crucial in the field of chemistry. It allows us to predict how a substance will react to changes in temperature when energy is added or removed. This experiment demonstrates how to measure and calculate the specific heat capacity of a given substance.
**Problem Statement:**
A chemist carefully measures the amount of heat needed to raise the temperature of a 0.53 kg sample of a pure substance from 10.7°C to 16.6°C. The experiment shows that 0.73 kJ of heat (Q) are needed. What can the chemist report for the specific heat capacity (c) of the substance? Make sure your answer has the correct number of significant digits.
**Given Data:**
- Mass of the substance (m): 0.53 kg
- Initial temperature (T1): 10.7°C
- Final temperature (T2): 16.6°C
- Heat added (Q): 0.73 kJ
**Objective:**
Calculate the specific heat capacity (c) of the substance using the given data.
**Formula:**
To determine the specific heat capacity, use the formula:
\[ c = \frac{Q}{m \cdot \Delta T} \]
where:
- \( Q \) is the amount of heat added (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( \Delta T \) is the change in temperature (in Kelvin/Celsius since the change is the same in both).
**Steps to Solve:**
1. Convert the heat added from kJ to J:
\[ 0.73 \, \text{kJ} = 0.73 \times 1000 \, \text{J} = 730 \, \text{J} \]
2. Calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T2 - T1 = 16.6°C - 10.7°C = 5.9°C \]
3. Substitute the values into the formula:
\[ c = \frac{730 \, \text{J}}{0.53 \, \text{kg} \times 5.9 \, \text{°C}} \]
4. Compute the result:
\[ c = \frac{730}{0
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