A 32.06 g sample of a substance is initially at 25.4 °C. After absorbing 1067 J of heat, the temperature of the substance is 132.2 °C. What is the specific heat (c) of the substance? J C = g. °C
A 32.06 g sample of a substance is initially at 25.4 °C. After absorbing 1067 J of heat, the temperature of the substance is 132.2 °C. What is the specific heat (c) of the substance? J C = g. °C
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:**
A 32.06 g sample of a substance is initially at 25.4 °C. After absorbing 1067 J of heat, the temperature of the substance is 132.2 °C. What is the specific heat (c) of the substance?
**Solution:**
To find the specific heat (c) of the substance, use the formula:
\[ c = \frac{q}{{m \cdot \Delta T}} \]
Where:
- \( q \) = heat absorbed (1067 J)
- \( m \) = mass of the sample (32.06 g)
- \( \Delta T \) = change in temperature (132.2 °C - 25.4 °C)
**Calculation:**
1. Calculate the change in temperature (\( \Delta T \)):
\[ \Delta T = 132.2 \, \text{°C} - 25.4 \, \text{°C} = 106.8 \, \text{°C} \]
2. Substitute the values into the formula:
\[ c = \frac{1067 \, \text{J}}{32.06 \, \text{g} \cdot 106.8 \, \text{°C}} \]
3. Simplify the expression to find \( c \):
\[ c = \ldots \, \text{(complete the calculation)} \]
The solution box is provided to enter the final specific heat value, expressed in units of J/g·°C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1cab08ae-8e5c-4fa2-a89b-725536e38ab3%2F0462ab28-392c-45a2-8ae4-aff6119b52aa%2F47ddfwj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 32.06 g sample of a substance is initially at 25.4 °C. After absorbing 1067 J of heat, the temperature of the substance is 132.2 °C. What is the specific heat (c) of the substance?
**Solution:**
To find the specific heat (c) of the substance, use the formula:
\[ c = \frac{q}{{m \cdot \Delta T}} \]
Where:
- \( q \) = heat absorbed (1067 J)
- \( m \) = mass of the sample (32.06 g)
- \( \Delta T \) = change in temperature (132.2 °C - 25.4 °C)
**Calculation:**
1. Calculate the change in temperature (\( \Delta T \)):
\[ \Delta T = 132.2 \, \text{°C} - 25.4 \, \text{°C} = 106.8 \, \text{°C} \]
2. Substitute the values into the formula:
\[ c = \frac{1067 \, \text{J}}{32.06 \, \text{g} \cdot 106.8 \, \text{°C}} \]
3. Simplify the expression to find \( c \):
\[ c = \ldots \, \text{(complete the calculation)} \]
The solution box is provided to enter the final specific heat value, expressed in units of J/g·°C.
Expert Solution
Step 1
Mass of sample = 32.06 g
Initial temperature (T1) = 25.4 °C
Final temperature (T2) = 132.2 °C
Heat absorb = 1067 J
c = specific heat
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