2. A lead pellet having a mass of 26.47 g at 89.98 °C was placed in a calorimeter containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet?
2. A lead pellet having a mass of 26.47 g at 89.98 °C was placed in a calorimeter containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet?
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
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![**Calorimetry Problem: Determining Specific Heat**
2. A lead pellet having a mass of 26.47 g at 89.98 °C was placed in a calorimeter containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet?
In this problem, concepts of calorimetry are used to find the specific heat capacity of a substance (lead in this case). To solve this, we use the principle that the heat lost by the lead is equal to the heat gained by the water.
To calculate the specific heat (\(c\)) of the lead pellet:
- **Mass of lead pellet (\(m\))**: 26.47 g
- **Initial temperature of lead**: 89.98 °C
- **Final temperature of water and lead**: 23.17 °C
- **Initial temperature of water**: 22.50 °C
- **Volume of water**: 100.0 mL (Assuming density of water is 1 g/mL, mass of water = 100.0 g)
The equation used:
\[ q_{\text{lead}} = m_{\text{lead}} \times c_{\text{lead}} \times \Delta T_{\text{lead}} \]
\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \]
And since \( q_{\text{lead}} = -q_{\text{water}} \):
\[ m_{\text{lead}} \times c_{\text{lead}} \times (T_{\text{final}} - T_{\text{initial, lead}}) = - m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}}) \]
Where:
- \( c_{\text{water}} \) is the specific heat capacity of water, typically 4.18 J/g°C.
- Solve for \( c_{\text{lead}} \).
This setup allows students to apply these relationships to find unknown specific heats.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96c43769-2331-4d91-b7f9-4a256a33e785%2F26f8408f-7aaf-4379-9a03-19b9faec1e4a%2Fa5v6ezo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calorimetry Problem: Determining Specific Heat**
2. A lead pellet having a mass of 26.47 g at 89.98 °C was placed in a calorimeter containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet?
In this problem, concepts of calorimetry are used to find the specific heat capacity of a substance (lead in this case). To solve this, we use the principle that the heat lost by the lead is equal to the heat gained by the water.
To calculate the specific heat (\(c\)) of the lead pellet:
- **Mass of lead pellet (\(m\))**: 26.47 g
- **Initial temperature of lead**: 89.98 °C
- **Final temperature of water and lead**: 23.17 °C
- **Initial temperature of water**: 22.50 °C
- **Volume of water**: 100.0 mL (Assuming density of water is 1 g/mL, mass of water = 100.0 g)
The equation used:
\[ q_{\text{lead}} = m_{\text{lead}} \times c_{\text{lead}} \times \Delta T_{\text{lead}} \]
\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \]
And since \( q_{\text{lead}} = -q_{\text{water}} \):
\[ m_{\text{lead}} \times c_{\text{lead}} \times (T_{\text{final}} - T_{\text{initial, lead}}) = - m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}}) \]
Where:
- \( c_{\text{water}} \) is the specific heat capacity of water, typically 4.18 J/g°C.
- Solve for \( c_{\text{lead}} \).
This setup allows students to apply these relationships to find unknown specific heats.
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