A 125-gg piece of metal is heated to 288 ∘C∘C and dropped into 85.0 gg of water at 12.0 ∘C∘C. The metal and water come to the same temperature of 24.0 ∘C∘C.

**Part A**

What is the specific heat, in J/g∘CJ/g∘C, of the metal?

Express your answer with the appropriate units.

To determine the specific heat (\(c\)) of the metal, we can use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water. 

### Step-by-step solution:

1. **Heat Transfer Equation:**
   The formula for heat transfer is given by:

   \[
   q = m \cdot c \cdot \Delta T
   \]

   Where:
   - \(q\) is the heat absorbed or released (in joules),
   - \(m\) is the mass (in grams),
   - \(c\) is the specific heat (in J/g°C),
   - \(\Delta T\) is the change in temperature (in °C).

2. **Given Data:**
   - Mass of the metal (\(m_{\text{metal}}\)): 125 g
   - Initial temperature of the metal (\(T_{\text{initial, metal}}\)): 288°C
   - Mass of the water (\(m_{\text{water}}\)): 85.0 g
   - Initial temperature of the water (\(T_{\text{initial, water}}\)): 12.0°C
   - Final equilibrium temperature (\(T_{\text{final}}\)): 24.0°C
   - Specific heat of water (\(c_{\text{water}}\)): 4.18 J/g°C

3. **Heat gained by water:**

   \[
   \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 24.0°C - 12.0°C = 12.0°C
   \]

   The heat gained by the water (\(q_{\text{water}}\)):

   \[
   q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
   \]

   \[
   q_{\text{water}} = 85.0 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 12.0°C
   \]

   \[
   q_{\text{water}} = 85.0 \cdot 4.18 \cdot 12.0
   \]

   \[
   q_{\text{water}} = 4264.8 \, \text{J}
   \]

4. **Heat lost by the metal:**

   \[
   \Delta T_{\text{metal}} = T_{\text{initial, metal}} - T_{\text{final}} = 288°C - 24.0°C = 264°C
   \]

   The heat lost by the metal (\(q_{\text{metal}}\)):

   \[
   q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
   \]

   Since the heat lost by the metal equals the heat gained by the water (by conservation of energy):

   \[
   q_{\text{metal}} = -q_{\text{water}}
   \]

   \[
   125 \cdot c_{\text{metal}} \cdot 264 = 4264.8
   \]

5. **Solve for the specific heat of the metal (\(c_{\text{metal}}\)):**

   \[
   c_{\text{metal}} = \frac{4264.8}{125 \cdot 264}
   \]

   \[
   c_{\text{metal}} = \frac{4264.8}{33000}
   \]

   \[
   c_{\text{metal}} \approx 0.129 \, \text{J/g°C}
   \]

### Answer:

The specific heat of the metal is approximately **0.129 J/g°C**.

---

Would you like more details on any part of the solution, or do you have any other questions?

### Related Questions:
1. What would be the final temperature if the mass of water were doubled?
2. How much heat would the metal release if its specific heat were 0.200 J/g°C?
3. If a different metal with a higher specific heat is used, how would the final temperature change?
4. How would the specific heat change if the initial temperature of water were higher?
5. Can the final temperature ever be higher than the initial temperature of the metal?

### Tip:
Remember that in heat exchange problems, the object losing heat will always cool down while the object gaining heat will warm up until they reach thermal equilibrium.

A 125-g piece of metal is heated to 288°C and dropped into 85.0 g of water at 12.0°C. The metal and water come to the same temperature of 24.0°C. What is the specific heat, in J/g°C, of the metal?

Learn how to calculate the specific heat of a metal using heat transfer equations and the conservation of energy principle. This problem involves determining the heat lost by the metal and the heat gained by water to find the specific heat capacity. Suitable for students in Grades 10-12.

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