A 1.3 g wafer of pure gold initially at 69.8 °C is submerged into 64.1 g of water at 26.9 °C in an insulated container. The specific heat capacity for gold is 0.128 J/g°C and the specific heat capacity for water is 4.18 J/g°C. What is the final temperature of the system when thermal equilibrium is reached?

• A. 28.0 °C
• B. 26.9 °C
• C. 27.0 °C
• D. 27.1 °C

Answer: (D)

Verified step by step guidance

1. Identify the principle of conservation of energy, which states that the heat lost by the gold will be equal to the heat gained by the water when they reach thermal equilibrium.
2. Write the equation for heat transfer: \( q_{gold} = -q_{water} \). This means the heat lost by gold is equal to the negative of the heat gained by water.
3. Use the formula for heat transfer: \( q = m \cdot c \cdot \Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.
4. Set up the equation using the given values: \( 1.3 \text{ g} \cdot 0.128 \text{ J/g°C} \cdot (T_{final} - 69.8°C) = -64.1 \text{ g} \cdot 4.18 \text{ J/g°C} \cdot (T_{final} - 26.9°C) \).
5. Solve the equation for \( T_{final} \) to find the final temperature when both substances reach thermal equilibrium.