A 32.8 g iron rod, initially at 22.8 °C, is submerged into an unknown mass of water at 64.0 °C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 °C. What is the mass of water?

• A. 12.5 g
• B. 25.0 g
• C. 75.0 g
• D. 50.0 g

Answer: (D)

Verified step by step guidance

1. Identify the principle of conservation of energy, which states that the heat lost by the iron rod will be equal to the heat gained by the water, as the system is insulated.
2. Use the formula for heat transfer: \( q = m \cdot c \cdot \Delta T \), where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
3. Calculate the heat lost by the iron rod. The specific heat capacity of iron is approximately 0.449 J/g°C. The change in temperature for the iron is \( 59.5 \text{°C} - 22.8 \text{°C} \). Substitute these values into the formula to find \( q_{\text{iron}} \).
4. Set up the equation for the heat gained by the water using the same formula. The specific heat capacity of water is 4.18 J/g°C. The change in temperature for the water is \( 59.5 \text{°C} - 64.0 \text{°C} \). Substitute these values into the formula to find \( q_{\text{water}} \).
5. Since the heat lost by the iron is equal to the heat gained by the water, set \( q_{\text{iron}} = q_{\text{water}} \) and solve for the mass of the water. This will give you the mass of the water in grams.