Part 1: Calculate the specific heat of Bronze using coffee cup calorimeter. Bronze Data Mass Of Metal (Bronze) Ttmetal) Mass Of Water Tie120) Tawater & metal) (5 pts) 19.9797 g 200 °C 150.00 g 25.00 °C 27.32 °C Calculations: You must show ALL work here. 1. Calculate qu20 Qn,0 = mH,0 x SH,0 × AT Qn;0 = 2. Calculate the specific heat of Bronze. Qn,0 = -qmetal = 4metal = mmetal x SHH,0 x ATmetal 9metal (massmetat X AT) (19.9797 g x (27.32 – 200.00)°C) SHmetal SHmetal g°C 3. % error Calculation (the true value for SHbronze is 0.435 g°C
Part 1: Calculate the specific heat of Bronze using coffee cup calorimeter. Bronze Data Mass Of Metal (Bronze) Ttmetal) Mass Of Water Tie120) Tawater & metal) (5 pts) 19.9797 g 200 °C 150.00 g 25.00 °C 27.32 °C Calculations: You must show ALL work here. 1. Calculate qu20 Qn,0 = mH,0 x SH,0 × AT Qn;0 = 2. Calculate the specific heat of Bronze. Qn,0 = -qmetal = 4metal = mmetal x SHH,0 x ATmetal 9metal (massmetat X AT) (19.9797 g x (27.32 – 200.00)°C) SHmetal SHmetal g°C 3. % error Calculation (the true value for SHbronze is 0.435 g°C
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Question
![**Title: Calculating the Specific Heat of Bronze Using a Coffee Cup Calorimeter**
**Introduction**
Understanding specific heat is crucial in material science and thermodynamics. In this tutorial, we will calculate the specific heat of bronze using a coffee cup calorimeter. Let's start with the necessary data and formulas, followed by detailed steps for calculations.
---
**Data Table: Bronze Data (5 pts)**
| Parameter | Value |
|--------------------------|----------------|
| Mass of Metal (Bronze) | 19.9797 g |
| Temperature of Metal (\(T_{\text{metal}}\)) | 200°C |
| Mass of Water | 150.00 g |
| Temperature of Water (\(T_{\text{H2O}}\)) | 25.00°C |
| Final Temperature (\(T_{\text{water & metal}}\)) | 27.32°C |
---
**Calculations**
*You must show all work here.*
1. **Calculate \(Q_{\text{H2O}}\)**
Formula:
\[
Q_{\text{H2O}} = m_{\text{H2O}} \times SH_{\text{H2O}} \times \Delta T
\]
Where \(Q_{\text{H2O}}\) is the heat absorbed by water, \(m_{\text{H2O}}\) is the mass of water, \(SH_{\text{H2O}}\) is the specific heat of water, and \(\Delta T\) is the change in temperature.
2. **Calculate the Specific Heat of Bronze**
The formula for the heat balance:
\[
Q_{\text{H2O}} = -Q_{\text{metal}}
\]
This implies:
\[
q_{\text{metal}} = m_{\text{metal}} \times SH_{\text{H2O}} \times \Delta T_{\text{metal}}
\]
*Specific Heat of Metal (\(SH_{\text{metal}}\)):*
\[
SH_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \times \Delta T} = \frac{19.9797 \, \text{g} \times (27.32 - 200.00)°C}{J/g°C}
\]
3. **%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3dd8c9ac-d087-46bb-b8c4-53ceffa446e7%2Fb96ce278-42db-47c5-9938-6e942eab131e%2F36tnek9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Specific Heat of Bronze Using a Coffee Cup Calorimeter**
**Introduction**
Understanding specific heat is crucial in material science and thermodynamics. In this tutorial, we will calculate the specific heat of bronze using a coffee cup calorimeter. Let's start with the necessary data and formulas, followed by detailed steps for calculations.
---
**Data Table: Bronze Data (5 pts)**
| Parameter | Value |
|--------------------------|----------------|
| Mass of Metal (Bronze) | 19.9797 g |
| Temperature of Metal (\(T_{\text{metal}}\)) | 200°C |
| Mass of Water | 150.00 g |
| Temperature of Water (\(T_{\text{H2O}}\)) | 25.00°C |
| Final Temperature (\(T_{\text{water & metal}}\)) | 27.32°C |
---
**Calculations**
*You must show all work here.*
1. **Calculate \(Q_{\text{H2O}}\)**
Formula:
\[
Q_{\text{H2O}} = m_{\text{H2O}} \times SH_{\text{H2O}} \times \Delta T
\]
Where \(Q_{\text{H2O}}\) is the heat absorbed by water, \(m_{\text{H2O}}\) is the mass of water, \(SH_{\text{H2O}}\) is the specific heat of water, and \(\Delta T\) is the change in temperature.
2. **Calculate the Specific Heat of Bronze**
The formula for the heat balance:
\[
Q_{\text{H2O}} = -Q_{\text{metal}}
\]
This implies:
\[
q_{\text{metal}} = m_{\text{metal}} \times SH_{\text{H2O}} \times \Delta T_{\text{metal}}
\]
*Specific Heat of Metal (\(SH_{\text{metal}}\)):*
\[
SH_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \times \Delta T} = \frac{19.9797 \, \text{g} \times (27.32 - 200.00)°C}{J/g°C}
\]
3. **%
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