We can solve this using the principle of heat transfer:

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

where:

$q = mc\Delta T$

• $m$ = mass (g)
• $c$ = specific heat capacity (J/g·°C)
• $\Delta T$ = change in temperature (°C)

Given data:

• Mass of metal: $m_{\text{metal}} = 122$ g
• Initial temperature of metal: $T_{\text{metal,initial}} = 292$°C
• Final temperature of both: $T_{\text{final}} = 57.5$°C
• Mass of water: $m_{\text{water}} = 84.5$ g
• Initial temperature of water: $T_{\text{water,initial}} = 26.0$°C
• Specific heat capacity of water: $c_{\text{water}} = 4.184$ J/g·°C

Step 1: Calculate heat gained by water

$q_{\text{water}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}$

$q_{\text{water}} = (84.5 \text{ g}) (4.184 \text{ J/g°C}) (57.5 - 26.0)$

$q_{\text{water}} = (84.5)(4.184)(31.5)$

$q_{\text{water}} = 11125.3 \text{ J}$

Since heat lost by the metal equals heat gained by the water:

$q_{\text{metal}} = -q_{\text{water}} = -11125.3 \text{ J}$

Step 2: Solve for $c_{\text{metal}}$

$q_{\text{metal}} = m_{\text{metal}} c_{\text{metal}} \Delta T_{\text{metal}}$

$-11125.3 = (122) c_{\text{metal}} (57.5 - 292)$

$-11125.3 = (122) c_{\text{metal}} (-234.5)$

$c_{\text{metal}} = \frac{11125.3}{(122)(234.5)}$

$c_{\text{metal}} = \frac{11125.3}{28689}$

$c_{\text{metal}} = 0.388 \text{ J/g°C}$

Final Answer:

$\mathbf{0.388} \text{ J/g°C}$

Would you like a step-by-step breakdown or further clarification?

Here are some related questions to explore:

1. How does specific heat capacity affect temperature change?
2. What are typical specific heat values for different metals?
3. How does mass influence heat transfer in this scenario?
4. What are the assumptions made in calorimetry problems like this?
5. How would the result change if heat loss to the environment were considered?

Tip: Specific heat capacity helps identify unknown metals based on their thermal properties!