1. A 50g sample of metal iron that was heated to 90 °C was dropped into a coffee cup calorimeter containing 200mL of water at room temperature of 25 °C. Applying the density of water (1g/mL) and specific heat capacity capacities of water and iron, determine the final temperature when the system reaches equilibrium.
1. A 50g sample of metal iron that was heated to 90 °C was dropped into a coffee cup calorimeter containing 200mL of water at room temperature of 25 °C. Applying the density of water (1g/mL) and specific heat capacity capacities of water and iron, determine the final temperature when the system reaches equilibrium.
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:**
A 50g sample of metal iron that was heated to 90 °C was dropped into a coffee cup calorimeter containing 200 mL of water at room temperature of 25 °C. Applying the density of water (1g/mL) and specific heat capacities of water and iron, determine the final temperature when the system reaches equilibrium.
**Explanation:**
To solve this problem, we need to apply the principle of calorimetry, where the heat lost by the iron will be gained by the water until thermal equilibrium is reached.
1. **Calculate the mass of water:** Since the density of water is 1g/mL, the mass is equal to the volume. Therefore, the mass of water is 200g.
2. **Use specific heat capacities:**
- Specific heat capacity of water: \( c_w = 4.18 \, \text{J/g°C} \)
- Specific heat capacity of iron: \( c_{Fe} = 0.45 \, \text{J/g°C} \)
3. **Set up the heat balance equation:**
\[
\text{Heat lost by iron} = \text{Heat gained by water}
\]
\[
m_{Fe} \cdot c_{Fe} \cdot (T_i - T_f) = m_w \cdot c_w \cdot (T_f - T_w)
\]
Where:
- \( m_{Fe} \) = mass of iron (50g)
- \( T_i \) = initial temperature of iron (90 °C)
- \( T_f \) = final temperature
- \( m_w \) = mass of water (200g)
- \( T_w \) = initial temperature of water (25 °C)
4. **Solve for \( T_f \) (final temperature):** Rearrange and solve the equation to find the equilibrium temperature.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F278fbf48-8175-4c44-ac0e-2265809c567f%2Fffc25c1d-b41e-4099-a281-f82bfbe3d1a1%2Fe51vac_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A 50g sample of metal iron that was heated to 90 °C was dropped into a coffee cup calorimeter containing 200 mL of water at room temperature of 25 °C. Applying the density of water (1g/mL) and specific heat capacities of water and iron, determine the final temperature when the system reaches equilibrium.
**Explanation:**
To solve this problem, we need to apply the principle of calorimetry, where the heat lost by the iron will be gained by the water until thermal equilibrium is reached.
1. **Calculate the mass of water:** Since the density of water is 1g/mL, the mass is equal to the volume. Therefore, the mass of water is 200g.
2. **Use specific heat capacities:**
- Specific heat capacity of water: \( c_w = 4.18 \, \text{J/g°C} \)
- Specific heat capacity of iron: \( c_{Fe} = 0.45 \, \text{J/g°C} \)
3. **Set up the heat balance equation:**
\[
\text{Heat lost by iron} = \text{Heat gained by water}
\]
\[
m_{Fe} \cdot c_{Fe} \cdot (T_i - T_f) = m_w \cdot c_w \cdot (T_f - T_w)
\]
Where:
- \( m_{Fe} \) = mass of iron (50g)
- \( T_i \) = initial temperature of iron (90 °C)
- \( T_f \) = final temperature
- \( m_w \) = mass of water (200g)
- \( T_w \) = initial temperature of water (25 °C)
4. **Solve for \( T_f \) (final temperature):** Rearrange and solve the equation to find the equilibrium temperature.
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