An 80.0 g sample of metal, initially at 96.0 °C, is placed into 150.0 g of water initially at 26.0 °C in a calorimeter. The final temperature of the water is 28.1 °C. What is the identity of the metal? (The specific heat of water is 4.18 J/g°C.)

• A. Iron (specific heat = 0.450 J/g°C)
• B. Copper (specific heat = 0.385 J/g°C)
• C. Aluminum (specific heat = 0.897 J/g°C)
• D. Lead (specific heat = 0.128 J/g°C)

Answer: (B)

Verified step by step guidance

1. Identify the principle of conservation of energy: The heat lost by the metal will be equal to the heat gained by the water. This can be expressed as: \( q_{metal} = -q_{water} \).
2. Calculate the heat gained by the water using the formula \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, \( m = 150.0 \text{ g} \), \( c = 4.18 \text{ J/g°C} \), and \( \Delta T = 28.1°C - 26.0°C \).
3. Set up the equation for the heat lost by the metal: \( q_{metal} = m_{metal} \cdot c_{metal} \cdot \Delta T_{metal} \), where \( m_{metal} = 80.0 \text{ g} \), \( \Delta T_{metal} = 28.1°C - 96.0°C \).
4. Since \( q_{metal} = -q_{water} \), substitute the expression for \( q_{water} \) into the equation for \( q_{metal} \) and solve for \( c_{metal} \), the specific heat capacity of the metal.
5. Compare the calculated specific heat capacity of the metal with the given options (Iron, Copper, Aluminum, Lead) to identify the metal.