When 20.0 g of water at 25 °C is mixed with 30.0 g of water at 80 °C in an insulated container, what is the final temperature of the mixture? (Assume no heat loss to the surroundings and that the specific heat capacity of water is 4.18 J/g·°C.)

• A. 80 °C
• B. 45 °C
• C. 52 °C
• D. 61 °C

Answer: (D)

Verified step by step guidance

1. Identify the principle of conservation of energy: the heat lost by the hot water will be equal to the heat gained by the cold water, since the container is insulated and no heat is lost to the surroundings.
2. Set up the heat transfer equations using the formula for heat: \(q = m \times c \times \Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.
3. Express the heat lost by the hot water as \(q_{hot} = m_{hot} \times c \times (T_{initial,hot} - T_{final})\) and the heat gained by the cold water as \(q_{cold} = m_{cold} \times c \times (T_{final} - T_{initial,cold})\).
4. Set the heat lost equal to the heat gained: \(m_{hot} \times c \times (T_{initial,hot} - T_{final}) = m_{cold} \times c \times (T_{final} - T_{initial,cold})\).
5. Solve the equation for the final temperature \(T_{final}\) by isolating it on one side and substituting the known values for masses, initial temperatures, and specific heat capacity.