Problem 97 - General Chemistry

Question

Suppose you mix \(21.0 \mathrm{~g}\) of water at \(52.7^{\circ} \mathrm{C}\) with \(54.9 \mathrm{~g}\) of water at \(31.5^{\circ} \mathrm{C}\) in an insulated cup. What is the maximum temperature of the solution after mixing?

Step-by-Step Solution

Answer

The maximum temperature after mixing is approximately 37.3°C.

Step 1: Understand Heat Transfer

When two bodies of water at different temperatures are mixed, heat is transferred from the warmer water to the cooler water until they reach thermal equilibrium. The total heat lost by the hot water is equal to the total heat gained by the cold water since the system is insulated.

Step 2: Write the Heat Transfer Equation

The heat lost or gained is given by the formula: \[ q = mc\Delta T \] where \( m \) is mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, \( c = 4.18 \text{ J/g°C}\).

Step 3: Set Up the Equation for Thermal Equilibrium

At thermal equilibrium, the absolute value of the heat lost by the hot water equals the heat gained by the cold water:\[ m_1c(T_{ ext{final}} - T_1) = m_2c(T_2 - T_{ ext{final}}) \]Substitute the values: \( m_1 = 21.0\, \text{g} \), \( T_1 = 52.7^{\circ}\text{C} \), \( m_2 = 54.9\, \text{g} \), and \( T_2 = 31.5^{\circ}\text{C} \).

Step 4: Simplify the Equation and Solve for Final Temperature

Both masses are in grams and the specific heat capacities cancel out, so:\[ 21.0(T_{ ext{final}} - 52.7) = 54.9(31.5 - T_{ ext{final}}) \]Expanding and simplifying:\[ 21.0T_{ ext{final}} - 1106.7 = 1727.85 - 54.9T_{ ext{final}} \]Combine like terms:\[ 21.0T_{ ext{final}} + 54.9T_{ ext{final}} = 1727.85 + 1106.7 \]\[ 75.9T_{ ext{final}} = 2834.55 \]Solving for \( T_{ ext{final}}:\)\[ T_{ ext{final}} = \frac{2834.55}{75.9} \approx 37.3^{\circ}\text{C} \]

Step 5: Verify the Solution

The final temperature should logically be between the initial temperatures (31.5°C and 52.7°C), and 37.3°C fits this requirement. Therefore, the calculations are consistent and correct.

Key Concepts

Heat Transfer

When you mix liquids at different temperatures, like the two quantities of water in the exercise, heat transfer naturally occurs. The warmer water transfers some of its heat energy to the cooler water. This process continues until both reach the same temperature. It's a fundamental aspect of thermodynamics. In our example, the warmer water at 52.7°C loses energy, while the cooler water at 31.5°C absorbs this energy. This movement of heat is happening without any physical contact beyond the water itself and doesn’t require an external medium, which is ideal in an insulated setting.

This energy exchange is guided by the principle that energy cannot be created or destroyed. Instead, it is conserved. Hence, the amount of heat lost by the hot water will exactly equal the amount of heat gained by the cooler water in an insulated system. This leads us to the concept of thermal equilibrium, where all parts of a system eventually reach a uniform temperature.

Thermal Equilibrium

Thermal equilibrium is a state reached when two substances in thermal contact no longer transfer heat between them. This means both substances settle at the same temperature. In the context of the exercise, after mixing the warmer and cooler water, thermal equilibrium is the point at which both have reached the same temperature, called the final temperature.

During this process, there is a continuous adjustment period where the hot water is cooling down while the cold water is heating up. The insulated cup in the problem prevents heat from entering or escaping, ensuring that the only heat exchange is between the two water bodies, maintaining the sum of their heat energy constant.

Reaching thermal equilibrium ensures that the heat exchange process has been completed. It's also a critical condition needed to accurately calculate the final temperature in the mixing problem.

Specific Heat Capacity

Specific heat capacity is a measure of how much energy is required to change the temperature of a substance. In scientific terms, it's the heat required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is approximately 4.18 J/g°C. This makes water relatively resistant to temperature changes.

Due to its high specific heat capacity, water can absorb or release a substantial amount of heat without undergoing significant temperature changes. In the exercise, the specific heat capacity allows us to calculate the amount of heat each portion of water can absorb or release during the mixing process. By knowing the mass of each water sample and the temperature change, we can precisely determine energy exchanges.

When applied to the heat transfer equation, specific heat capacity helps to equate the heat gained by the cold water to the heat lost by the hot water, leading us toward the calculation of the final equilibrium temperature.

Final Temperature Calculation

Calculating the final temperature in a mixing problem involves setting up a heat balance equation based on heat transfer principles. In this exercise, the equation reflects that the total heat lost by the warmer water equals the total heat gained by the cooler water. This can be written mathematically using the formula: \[ q = mc\Delta T \]where:

• \( m \) is the mass of water.
• \( c \) is the specific heat capacity.
• \( \Delta T \) is the change in temperature.

Both portions of water share the same specific heat capacity, allowing it to cancel out in the calculations. This simplifies the problem, reducing it to the equation:\[ m_1(T_{final} - T_1) = m_2(T_2 - T_{final}) \]By substituting known values, we solve for the final temperature \( T_{final} \), which accurately computes to approximately 37.3°C. This result confirms that the water temperature is indeed between the starting temperatures of 31.5°C and 52.7°C. Understanding these calculations is crucial for predicting outcomes in thermal systems.