A 2.0 kg metal weight at 80 °C is placed into 1.0 kg of water at 20 °C in an insulated container. Assuming no heat is lost to the surroundings and the specific heat capacity of the metal is 0.45 J/g·°C and that of water is 4.18 J/g·°C, what is the final temperature of both the weight and the water at thermal equilibrium?

• A. Approximately 24 °C
• B. Approximately 32 °C
• C. Approximately 50 °C
• D. Approximately 36 °C

Answer: (D)

Verified step by step guidance

1. Convert the masses of the metal and water from kilograms to grams because the specific heat capacities are given in J/g·°C. Use the conversion: \$1\, \text{kg} = 1000\, \text{g}\(. So, \)m_{metal} = 2000\, \text{g}\( and \)m_{water} = 1000\, \text{g}$.
2. Set up the heat transfer equation based on the principle of conservation of energy: the heat lost by the metal equals the heat gained by the water. Mathematically, this is \(q_{metal} + q_{water} = 0\) or \(q_{metal} = -q_{water}\).
3. Express the heat transfer for each substance using the formula \(q = m \times c \times \Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. For the metal: \(q_{metal} = m_{metal} \times c_{metal} \times (T_{final} - T_{initial, metal})\). For the water: \(q_{water} = m_{water} \times c_{water} \times (T_{final} - T_{initial, water})\).
4. Substitute the known values and set up the equation: \(m_{metal} \times c_{metal} \times (T_{final} - 80) = - m_{water} \times c_{water} \times (T_{final} - 20)\), where \(T_{final}\) is the unknown final temperature.
5. Solve the equation algebraically for \(T_{final}\) by expanding both sides, collecting like terms, and isolating \(T_{final}\). This will give the final equilibrium temperature of both the metal and the water.