We can use the principle of conservation of energy, which states that the heat lost by the metal is equal to the heat gained by the water:

$q_{\text{metal}} = -q_{\text{water}}$

Step 1: Use the Heat Transfer Formula

The heat transfer equation is:

$q = mc\Delta T$

where:

• $q$ is the heat energy (J),
• $m$ is the mass (g),
• $c$ is the specific heat capacity (J/g·°C),
• $\Delta T$ is the temperature change (°C).

Step 2: Calculate the Heat Gained by Water

The specific heat of water is $c_{\text{water}} = 4.18$ J/g·°C. The temperature change for water is:

$\Delta T_{\text{water}} = 57.5^\circ C - 26.0^\circ C = 31.5^\circ C$

$q_{\text{water}} = (84.5 \text{ g}) (4.18 \text{ J/g°C}) (31.5^\circ C)$

$q_{\text{water}} = 11118.885 \text{ J}$

Step 3: Calculate the Specific Heat of the Metal

The temperature change for the metal is:

$\Delta T_{\text{metal}} = 57.5^\circ C - 292^\circ C = -234.5^\circ C$

Since the metal loses heat:

$q_{\text{metal}} = -q_{\text{water}}$

$(122 \text{ g}) (c_{\text{metal}}) (-234.5^\circ C) = -11118.885 \text{ J}$

Solving for $c_{\text{metal}}$:

$c_{\text{metal}} = \frac{11118.885}{122 \times 234.5}$

$c_{\text{metal}} \approx 0.39 \text{ J/g°C}$

Final Answer:

$c_{\text{metal}} \approx 0.39 \text{ J/g°C}$

This value suggests the metal could be copper or a similar material.

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Do you want a more detailed breakdown?

Related Questions:

1. What are some common metals with a specific heat near 0.39 J/g°C?
2. How does the specific heat of a metal relate to its atomic structure?
3. What would happen if a different liquid (e.g., oil) was used instead of water?
4. How does the conservation of energy apply to heat transfer problems?
5. How would the final temperature change if the mass of water was increased?

Tip:

Always make sure the units are consistent when solving heat transfer problems!