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.

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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.
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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