To find the equilibrium temperature of the water and quartz, we can use the principle of conservation of energy. Since no heat is lost to the surroundings, the heat lost by the water will equal the heat gained by the quartz.

Given:

• Mass of quartz ($m_q$) = 20.0 g
• Specific heat capacity of quartz ($c_q$) = 0.730 J/g°C
• Initial temperature of quartz ($T_{q,i}$) = 4.4°C
• Mass of water ($m_w$) = 250.0 g
• Specific heat capacity of water ($c_w$) = 4.184 J/g°C
• Initial temperature of water ($T_{w,i}$) = 60.0°C
• Final equilibrium temperature = $T_f$ (unknown)

Step 1: Set up the heat transfer equation.

The heat gained by the quartz is: $q_{\text{quartz}} = m_q \cdot c_q \cdot (T_f - T_{q,i})$ The heat lost by the water is: $q_{\text{water}} = m_w \cdot c_w \cdot (T_f - T_{w,i})$

Since no heat is lost to the surroundings, the heat gained by the quartz equals the heat lost by the water: $m_q \cdot c_q \cdot (T_f - T_{q,i}) = - m_w \cdot c_w \cdot (T_f - T_{w,i})$

Step 2: Substitute values into the equation.

$20.0 \cdot 0.730 \cdot (T_f - 4.4) = - 250.0 \cdot 4.184 \cdot (T_f - 60.0)$

Step 3: Solve for $T_f$.

First, simplify both sides of the equation: $14.6 \cdot (T_f - 4.4) = -1046.0 \cdot (T_f - 60.0)$

Expand both sides: $14.6 T_f - 64.24 = -1046.0 T_f + 62760$

Now, combine like terms: $14.6 T_f + 1046.0 T_f = 62760 + 64.24$ $1060.6 T_f = 62824.24$

Solve for $T_f$: $T_f = \frac{62824.24}{1060.6} \approx 59.2^\circ C$

Final Answer:

The equilibrium temperature of the water is approximately 59.2°C.

Let me know if you'd like further details on the calculation! Here are some related questions:

1. How does the specific heat capacity of materials affect the final equilibrium temperature?
2. What would happen if the specific heat of quartz were much higher or lower?
3. How would changing the mass of water alter the final temperature?
4. What assumptions are made in this heat exchange process?
5. What is the significance of ignoring heat exchange with the container or surroundings?

Tip: Always double-check units when performing heat transfer calculations, especially when dealing with different materials.