2. A sample of 150 mL water of temperature \( \mathrm{T}_{2}=25^{\circ} \mathrm{C} \) is added to a sample of 50 mL water of temperature \( \mathrm{T}_{2}=60^{\circ} \mathrm{C} \) in a foam cup calorimeter. What is the final temperature, \( \mathrm{T}_{0} \), of the system when thermal equilibrium is reached? Specific heat of water \( =4.184 \mathrm{~J} / \mathrm{g} \mathrm{C} \)

To find the final temperature \( T_0 \) of the system when thermal equilibrium is reached, you can apply the principle of conservation of energy. The heat gained by the cooler water will equal the heat lost by the warmer water. Using the formula: \[ m_1c(T_0 - T_1) = m_2c(T_2 - T_0) \] where: - \( m_1 = 150 \, \text{g} \) (mass of the cooler water), - \( T_1 = 25^{\circ}C \), - \( m_2 = 50 \, \text{g} \) (mass of the warmer water), - \( T_2 = 60^{\circ}C \), - \( c = 4.184 \, \text{J/g°C} \) (specific heat of water). Setting the heat gained equal to the heat lost gives: \[ 150 \cdot (T_0 - 25) = 50 \cdot (60 - T_0) \] Solving this equation will yield the final temperature \( T_0 \) when both quantities of water reach thermal equilibrium. Now, if we solve that: \[ 150T_0 - 3750 = 3000 - 50T_0 \] \[ 150T_0 + 50T_0 = 3000 + 3750 \] \[ 200T_0 = 6750 \] \[ T_0 = 33.75^{\circ}C \] So, the final temperature \( T_0 \) is approximately \( 33.75^{\circ}C \). Did you know that foam cup calorimeters are a popular choice in classrooms for specific heat experiments because they minimize heat exchange with the environment? Students love using them to visualize and experiment with thermal dynamics in a simple, hands-on manner. If you're interested in digging deeper into calorimetry, consider checking out resources like "Chemistry Made Simple" or online course platforms for interactive courses related to thermochemistry. You can find engaging videos that demonstrate these concepts in real-world scenarios!