Problem 14 - Chemistry and Chemical Reactivity

Question

When \(108 \mathrm{g}\) of water at a temperature of \(22.5^{\circ} \mathrm{C}\) is mixed with \(65.1 \mathrm{g}\) of water at an unknown temperature, the final temperature of the resulting mixture is \(47.9^{\circ} \mathrm{C} .\) What was the initial temperature of the second sample of water?

Step-by-Step Solution

Answer

The initial temperature of the second water sample was \(90.3^{\circ}C\).

Step 1: Understanding Heat Transfer

When two bodies of water are mixed, heat is transferred from the hotter water to the cooler water until they reach a final equilibrium temperature. The heat lost by the hot water is equal to the heat gained by the cold water: \(m_1c(T_f - T_1) = m_2c(T_2 - T_f)\), where \(m_1\), \(m_2\) are the masses and \(T_1\), \(T_2\) and \(T_f\) are the initial and final temperatures. \(c\) is the specific heat of water, which is the same and cancels out.

Step 2: Substitute Known Values

Substitute the known values into the equation. Here, \(m_1 = 108\,\mathrm{g}\), \(T_1 = 22.5^{\circ}\mathrm{C}\), \(m_2 = 65.1\,\mathrm{g}\), and \(T_f = 47.9^{\circ}\mathrm{C}\). The equation becomes: \( 108(T_f - 22.5) = 65.1(T_x - T_f)\).

Step 3: Solve for \(T_x\)

First, calculate \(108(47.9 - 22.5)\) and \(65.1(T_x - 47.9)\). So, solve \( 108 \times 25.4 = 65.1T_x - 3120.69\). Calculate the left-side as \(2743.2 = 65.1T_x - 3120.69\), then reorganize to find \(T_x\): \( 65.1T_x = 2743.2 + 3120.69\), which results in \(T_x = \frac{5879.89}{65.1}\).

Step 4: Compute the Temperature

Finally, compute \(T_x = \frac{5879.89}{65.1} = 90.3^{\circ}C\). Hence, the initial temperature of the second sample of water was \(90.3^{\circ}C\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in heat transfer. It explains how much heat energy is needed to change the temperature of a certain substance. - Every substance has its unique specific heat capacity, which tells us how resistant it is to temperature change when absorbing or losing heat. - For water, this specific heat capacity is relatively high at 4.18 J/g°C. In this problem, specific heat capacity helps us understand how heat transfer involves the balanced energy requirements of different quantities of water when their temperatures change. These values were canceled out in our calculations because the specific heat of water is the same on both sides of the equation.

Equilibrium Temperature

Equilibrium temperature is the point at which two or more bodies in thermal contact reach the same temperature. It is achieved when the heat lost by the hotter object is equal to the heat gained by the cooler one. - For instance, when warmer and cooler water samples are mixed, they exchange heat energy until a uniform temperature is obtained. - In this exercise, it was 47.9°C — both water samples stabilize at this temperature. Calculating this equilibrium temperature involves considering the mass and initial temperatures of the mixed water bodies. It reveals how energy is conserved during heat transfer.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It governs how energy exchanges occur in systems, providing the framework for heat transfer analysis. - In our problem, the principles of thermodynamics imply that energy is conserved when two water bodies with different temperatures interact. - The law of energy conservation ensures that the heat the hot water loses equals the heat gained by the cold sample, bringing forth equilibrium. Understanding these interactions is critical, allowing us to derive equations that predict temperature changes, as demonstrated in the solution process.

Chemical Calculations

Chemical calculations often build on thermodynamic principles to solve problems like those involving heat transfer. They require mathematical steps to determine unknown variables, like the initial temperature of a water sample. - In this exercise, calculations consist of substituting known values into heat transfer equations and solving for unknowns, such as rearranging terms to isolate the temperature variable. - These calculations rely heavily on algebra to evaluate the outcome, as seen when calculating the unknown initial temperature of the second water sample. These steps make complex thermodynamic solutions accessible by breaking down the process into manageable calculations.