Problem 98 - General Chemistry

Question

Suppose you mix \(20.5 \mathrm{~g}\) of water at \(66.2^{\circ} \mathrm{C}\) with \(45.4 \mathrm{~g}\) of water at \(35.7^{\circ} \mathrm{C}\) in an insulated cup. What is the maximum temperature of the solution after mixing?

Step-by-Step Solution

Answer

The final temperature is approximately \(45.1^{\circ} \mathrm{C}\).

Step 1: Define the Problem

We need to find the final equilibrium temperature of two portions of water at different temperatures mixed together in an insulated container. This temperature is achieved when the heat lost by the hotter water equals the heat gained by the cooler water.

Step 2: List the Known Data

The mass of the first water portion is \(20.5 \mathrm{~g}\) with an initial temperature of \(66.2^{\circ} \mathrm{C}\). The mass of the second water portion is \(45.4 \mathrm{~g}\) with an initial temperature of \(35.7^{\circ} \mathrm{C}\). The specific heat capacity of water is \(4.18 \mathrm{~J/g^{\circ}C}\).

Step 3: Set up the Heat Transfer Equation

Since no heat is lost to the surroundings, the heat lost by the hot water will equal the heat gained by the cold water. This can be written as: \( m_1c(T_f - T_1) = m_2c(T_2 - T_f) \)where \( m_1 = 20.5 \mathrm{~g} \), \( T_1 = 66.2^{\circ} \mathrm{C} \), \( m_2 = 45.4 \mathrm{~g} \), and \( T_2 = 35.7^{\circ} \mathrm{C} \). \( c \) is the specific heat capacity.

Step 4: Simplify the Equation

Since \( c \) is common on both sides, it can be canceled out. This simplifies to:\(20.5(T_f - 66.2) = 45.4(35.7 - T_f)\)

Step 5: Solve for Final Temperature \(T_f\)

Expand both sides of the equation:\( 20.5T_f - 1355.1 = 1620.78 - 45.4T_f \)Rearrange terms to isolate \(T_f\):\( 20.5T_f + 45.4T_f = 1620.78 + 1355.1 \)\( 65.9T_f = 2975.88 \)Divide both sides to solve for \(T_f\):\( T_f = \frac{2975.88}{65.9} \approx 45.1^{\circ} \mathrm{C} \)

Step 6: Conclusion

The maximum temperature reached by the mixture, assuming no heat is lost to the surroundings, is approximately \(45.1^{\circ} \mathrm{C}\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property that tells us how much heat is needed to change the temperature of a given mass of a substance by one degree Celsius. It is expressed in units of joules per gram per degree Celsius (J/g°C). The formula to calculate heat change involves the specific heat capacity, and it is given by:

• Heat change (\( q \)) = mass (\( m \)) × specific heat capacity (\( c \)) × change in temperature (\( \Delta T \))

Water, being a common substance, has a specific heat capacity of 4.18 J/g°C. This means it takes 4.18 joules to raise the temperature of 1 gram of water by 1°C. This property is crucial for solving problems related to heat transfer because it allows us to calculate how much heat is absorbed or released as a substance's temperature changes.
The specific heat capacity remains constant unless the substance undergoes a phase change, such as melting or boiling. In this context, knowing the specific heat allows us to predict how much temperature change will occur when two masses of water at different temperatures are mixed.

Thermal Equilibrium

Thermal equilibrium occurs when two objects in contact do not exchange heat with each other. In simpler terms, it is when they reach the same temperature. It is a fundamental concept in thermodynamics which states that when two bodies are in thermal contact, heat will flow from the hotter object to the cooler one until their temperatures equalize.

• In the problem, two masses of water at different temperatures are mixed, and they eventually reach the same final temperature, known as the equilibrium temperature.
• This equilibrium temperature is the state where the heat lost by the warmer water equals the heat gained by the cooler water.

This concept is important because thermal equilibrium is the goal in practical heat transfer applications, such as in designing home heating systems or cooling devices. By understanding thermal equilibrium, one can predict how long it takes for materials to reach the same temperature and how much energy is required for the process to occur.

Conservation of Energy

The principle of conservation of energy is one of the cornerstones of physics, stating that energy cannot be created or destroyed, only transferred or converted from one form to another. When applying this principle to heat transfer, it implies that the total amount of heat energy in an isolated system remains constant. This means any heat lost by one object is gained by another.

• In the context of the exercise, the conservation of energy principle means that the heat lost by the hot water is completely absorbed by the cold water.
• This is expressed mathematically by the equation: \( m_1c(T_f - T_1) = m_2c(T_2 - T_f) \).

Here, the heat lost by the hotter water (\( m_1c(T_f - T_1) \)) must be equal to the heat gained by the cooler water (\( m_2c(T_2 - T_f) \)).
This concept ensures that the total energy before mixing equals the total energy after reaching thermal equilibrium. Understanding conservation of energy allows us to solve complex problems involving multiple energy transfers and transformations in a straightforward and predictable way.