A 90.0 g piece of metal, initially at 98.60 °C, is placed into 120.0 g of water initially at 24.30 °C. If the final temperature of the water is 34.00 °C, what is the specific heat capacity of the metal? (The specific heat of water is 4.184 J/g·°C).

• A. 0.500 J/g·°C
• B. 0.215 J/g·°C
• C. 0.750 J/g·°C
• D. 0.385 J/g·°C

Answer: (D)

Verified step by step guidance

1. Identify the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water.
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 gained by the water using its mass (120.0 g), specific heat capacity (4.184 J/g·°C), and temperature change (34.00 °C - 24.30 °C).
4. Set the heat lost by the metal equal to the heat gained by the water. Use the mass of the metal (90.0 g), its unknown specific heat capacity \( c \), and its temperature change (98.60 °C - 34.00 °C) to express the heat lost by the metal.
5. Solve the equation for the specific heat capacity \( c \) of the metal by equating the heat lost by the metal to the heat gained by the water and isolating \( c \).