A 138 g quantity of a certain metal, initially at 125 °C, is dropped into an insulated foam beaker containing 81.0 g of water at 20.0 °C.  The final temperature of the metal and water in the beaker is measured at 40.0 °C.  Assume that the heat flowing into the foam beaker can be ignored (heat is only exchanged between the sample and the water).  What is the specific heat capacity of the metal?

To find the specific heat capacity (\(c\)) of the metal, we can apply the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the water since no heat is lost to the surroundings.

### Step-by-Step Solution:

1. **Heat gained by the water**:
   The heat gained by the water is given by the formula:

   \[
   q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
   \]
   Where:
   - \(m_{\text{water}}\) = mass of water = 81.0 g
   - \(c_{\text{water}}\) = specific heat capacity of water = \(4.184 \, \text{J/g°C}\)
   - \(\Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial}} = 40.0°C - 20.0°C = 20.0°C\)

   So,

   \[
   q_{\text{water}} = 81.0 \, \text{g} \cdot 4.184 \, \text{J/g°C} \cdot 20.0 \, \text{°C}
   \]
   \[
   q_{\text{water}} = 6781.92 \, \text{J}
   \]

2. **Heat lost by the metal**:
   The heat lost by the metal is:

   \[
   q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
   \]
   Where:
   - \(m_{\text{metal}}\) = mass of metal = 138 g
   - \(c_{\text{metal}}\) = specific heat capacity of the metal (this is what we want to find)
   - \(\Delta T_{\text{metal}} = T_{\text{initial}} - T_{\text{final}} = 125.0°C - 40.0°C = 85.0°C\)

   So,

   \[
   q_{\text{metal}} = 138 \, \text{g} \cdot c_{\text{metal}} \cdot 85.0 \, \text{°C}
   \]

3. **Setting heat lost by metal equal to heat gained by water**:

   Since \( q_{\text{metal}} = q_{\text{water}} \), we can write:

   \[
   138 \, \text{g} \cdot c_{\text{metal}} \cdot 85.0 \, \text{°C} = 6781.92 \, \text{J}
   \]

4. **Solving for \( c_{\text{metal}} \)**:

   \[
   c_{\text{metal}} = \frac{6781.92 \, \text{J}}{138 \, \text{g} \cdot 85.0 \, \text{°C}}
   \]
   \[
   c_{\text{metal}} = \frac{6781.92}{11730} \, \text{J/g°C}
   \]
   \[
   c_{\text{metal}} \approx 0.578 \, \text{J/g°C}
   \]

### Final Answer:
The specific heat capacity of the metal is approximately **0.578 J/g°C**.

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Do you need further details on any of the steps? 

Here are five related questions:
1. How does the concept of thermal equilibrium apply in this scenario?
2. What assumptions are made about the insulation and heat loss in this problem?
3. How would the result change if the foam beaker absorbed some heat?
4. What is the significance of specific heat capacity in material science?
5. How would the final temperature change if the mass of water were different?

**Tip:** Always ensure that you are working with consistent units (grams, joules, °C) when solving thermodynamics problems.

A 138 g quantity of a certain metal, initially at 125 °C, is dropped into an insulated foam beaker containing 81.0 g of water at 20.0 °C. The final temperature of the metal and water in the beaker is measured at 40.0 °C. Assume that the heat flowing into the foam beaker can be ignored (heat is only exchanged between the sample and the water). What is the specific heat capacity of the metal?

Explore a detailed solution to calculate the specific heat capacity of a 138 g metal sample dropped into water. This thermodynamics problem demonstrates the principle of conservation of energy, using heat transfer formulas to find the specific heat. Suitable for Grades 11-12 and college-level students.

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