Questions: 10.0 g Cu was heated to 600°C then dropped into water at 25°C If gow hod 325 g water what is the final temp. Cω=0.385 S / g°C C4.0=4.184 J / g°C

Solution Steps

We are given the following information:

• Mass of copper, $m_{\text{Cu}} = 10.0 \, \text{g}$
• Initial temperature of copper, $T_{\text{Cu, initial}} = 600^\circ \text{C}$
• Mass of water, $m_{\text{water}} = 325 \, \text{g}$
• Initial temperature of water, $T_{\text{water, initial}} = 25^\circ \text{C}$
• Specific heat capacity of copper, $C_{\text{Cu}} = 0.385 \, \text{J/g}^\circ \text{C}$
• Specific heat capacity of water, $C_{\text{water}} = 4.184 \, \text{J/g}^\circ \text{C}$

When the copper is dropped into the water, heat will transfer from the copper to the water until thermal equilibrium is reached. The heat lost by the copper will equal the heat gained by the water:

$$m_{\text{Cu}} \cdot C_{\text{Cu}} \cdot (T_{\text{final}} - T_{\text{Cu, initial}}) = -m_{\text{water}} \cdot C_{\text{water}} \cdot (T_{\text{final}} - T_{\text{water, initial}})$$

Substitute the known values into the equation:

$$10.0 \cdot 0.385 \cdot (T_{\text{final}} - 600) = -325 \cdot 4.184 \cdot (T_{\text{final}} - 25)$$

Simplify and solve for $T_{\text{final}}$:

$$3.85 \cdot (T_{\text{final}} - 600) = -1358.8 \cdot (T_{\text{final}} - 25)$$

Expanding both sides:

$$3.85T_{\text{final}} - 2310 = -1358.8T_{\text{final}} + 33970$$

Combine like terms:

$$3.85T_{\text{final}} + 1358.8T_{\text{final}} = 33970 + 2310$$

$$1362.65T_{\text{final}} = 36280$$

Solve for $T_{\text{final}}$:

$$T_{\text{final}} = \frac{36280}{1362.65} \approx 26.625^\circ \text{C}$$

Final Answer