A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('heavy water,' specific heat capacity = 4.211 J/°C g) at 25.5°C. The final temperature of the platinum and deuterium oxide mixture is 29.1°C. The density of platinum is 21.45 g/ cm³ and the density of deuterium oxide is 1.11 g/mL. What is the edge length of the cube of platinum, in centimeters?

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### Heat Transfer and Density Calculations

A cubic piece of platinum metal, with a specific heat capacity of 0.1256 J/°C·g, is initially at a temperature of 200.0°C. When submerged into 1.00 L of deuterium oxide ('heavy water'), which has a specific heat capacity of 4.211 J/°C·g, the system's initial temperature is 25.5°C. Upon reaching thermal equilibrium, the final temperature of the platinum and deuterium oxide mixture is 29.1°C.

#### Given Data

- **Platinum Metal:**
  - Initial Temperature = 200.0°C
  - Specific Heat Capacity = 0.1256 J/°C·g
  - Density = 21.45 g/cm³

- **Deuterium Oxide:**
  - Volume = 1.00 L
  - Initial Temperature = 25.5°C
  - Specific Heat Capacity = 4.211 J/°C·g
  - Density = 1.11 g/mL

#### Problem

Calculate the edge length of the cube of platinum, expressed in centimeters.

### Calculation Approach

1. **Calculate the Mass of Deuterium Oxide:**
   \[
   \text{Mass} = \text{Density} \times \text{Volume} = 1.11 \, \text{g/mL} \times 1000 \, \text{mL} = 1110 \, \text{g}
   \]

2. **Energy Transfer (Heat Balance):**
   The heat lost by platinum equals the heat gained by deuterium oxide:
   \[
   (m_{\text{Pt}} \cdot c_{\text{Pt}} \cdot \Delta T_{\text{Pt}}) = (m_{\text{D2O}} \cdot c_{\text{D2O}} \cdot \Delta T_{\text{D2O}})
   \]
   where:
   - \( \Delta T_{\text{Pt}} = 200.0°C - 29.1°C \)
   - \( \Delta T_{\text{D2O}} = 29.1°C - 25.5°C \)

3. **Calculate the Mass of Platinum:**
   Use the formula above to find \( m_{\text{Pt}} \
Transcribed Image Text:### Heat Transfer and Density Calculations A cubic piece of platinum metal, with a specific heat capacity of 0.1256 J/°C·g, is initially at a temperature of 200.0°C. When submerged into 1.00 L of deuterium oxide ('heavy water'), which has a specific heat capacity of 4.211 J/°C·g, the system's initial temperature is 25.5°C. Upon reaching thermal equilibrium, the final temperature of the platinum and deuterium oxide mixture is 29.1°C. #### Given Data - **Platinum Metal:** - Initial Temperature = 200.0°C - Specific Heat Capacity = 0.1256 J/°C·g - Density = 21.45 g/cm³ - **Deuterium Oxide:** - Volume = 1.00 L - Initial Temperature = 25.5°C - Specific Heat Capacity = 4.211 J/°C·g - Density = 1.11 g/mL #### Problem Calculate the edge length of the cube of platinum, expressed in centimeters. ### Calculation Approach 1. **Calculate the Mass of Deuterium Oxide:** \[ \text{Mass} = \text{Density} \times \text{Volume} = 1.11 \, \text{g/mL} \times 1000 \, \text{mL} = 1110 \, \text{g} \] 2. **Energy Transfer (Heat Balance):** The heat lost by platinum equals the heat gained by deuterium oxide: \[ (m_{\text{Pt}} \cdot c_{\text{Pt}} \cdot \Delta T_{\text{Pt}}) = (m_{\text{D2O}} \cdot c_{\text{D2O}} \cdot \Delta T_{\text{D2O}}) \] where: - \( \Delta T_{\text{Pt}} = 200.0°C - 29.1°C \) - \( \Delta T_{\text{D2O}} = 29.1°C - 25.5°C \) 3. **Calculate the Mass of Platinum:** Use the formula above to find \( m_{\text{Pt}} \
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