Question

A 0.0675 kg ingot of metal is heated to 242◦C and then is dropped into a beaker containing 0.374 kg of water initially at 23◦C. If the final equilibrium state of the mixed system is 25.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg ·◦ C. Answer in units of J/kg ·◦ C.

Answer 1

To find the specific heat of the metal, we can use the principle of conservation of energy, which states that the heat lost by the metal when it cools down will be equal to the heat gained by the water as it warms up.

1. **Define the variables:**
- Mass of the metal, \( m_m = 0.0675 \) kg
- Initial temperature of the metal, \( T_{m,initial} = 242 \) °C
- Final temperature of the system, \( T_{final} = 25.4 \) °C
- Mass of the water, \( m_w = 0.374 \) kg
- Initial temperature of the water, \( T_{w,initial} = 23 \) °C
- Specific heat of water, \( c_w = 4186 \) J/kg·°C
- Specific heat of the metal, \( c_m = ? \) J/kg·°C

2. **Set up the heat transfer equations:**
The heat lost by the metal will be equal to the heat gained by the water:

\[
m_m c_m (T_{m,initial} - T_{final}) = m_w c_w (T_{final} - T_{w,initial})
\]

3. **Substitute the known values:**
\[
0.0675 \, c_m (242 - 25.4) = 0.374 \, 4186 \, (25.4 - 23)
\]

4. **Calculate the temperature differences:**
- For metal: \( 242 - 25.4 = 216.6 \) °C
- For water: \( 25.4 - 23 = 2.4 \) °C

5. **Rewrite the equation with these values:**
\[
0.0675 \, c_m \, (216.6) = 0.374 \, 4186 \, (2.4)
\]

6. **Calculate the right hand side:**
\[
0.374 \times 4186 \times 2.4 \approx 3,139.76 \, \text{J}
\]

7. **Substituting this value into the equation:**
\[
0.0675 \, c_m \, (216.6) = 3139.76
\]

8. **Solving for \( c_m \):**
\[
c_m = \frac{3139.76}{0.0675 \times 216.6}
\]

- Calculate the denominator:
\[
0.0675 \times 216.6 \approx 14.6205
\]
- Then calculate \( c_m \):
\[
c_m = \frac{3139.76}{14.6205} \approx 214.55 \, \text{J/kg} \cdot \text{°C}
\]

The specific heat of the metal is approximately \( \boxed{214.55} \) J/kg·°C.