Problem 115 - Chemistry: An Atoms-Focused Approach

Question

An insulated container holds \(50.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} .\) A \(7.25 \mathrm{g}\) sample of copper that had been heated to \(100.1^{\circ} \mathrm{C}\) is dropped into the water. What is the final shared temperature of the copper and the water?

Step-by-Step Solution

Answer

Answer: The final shared temperature of the copper and water in the insulated container is approximately 28.3°C.

Step 1: Write down the given values

We have: - Mass of water (m1) = 50 g - Initial temperature of water (T1_initial) = 25°C - Specific heat capacity of water (c1) = 4.18 J/g°C - Mass of copper (m2) = 7.25 g - Initial temperature of copper (T2_initial) = 100.1°C - Specific heat capacity of copper (c2) = 0.386 J/g°C

Step 2: Calculate the heat transfer for water

We will use the formula Q = mcΔT. We will first calculate the heat transfer for water (Q1). Let the final temperature of the system be Tf. Q1 = m1 * c1 * (Tf - T1_initial)

Step 3: Calculate the heat transfer for copper

Now, we will calculate the heat transfer for copper (Q2). Q2 = m2 * c2 * (T2_initial - Tf)

Step 4: Equate the heat transfer of water and copper

Since the heat lost by copper is equal to the heat gained by water, we can set Q1 = Q2. m1 * c1 * (Tf - T1_initial) = m2 * c2 * (T2_initial - Tf)

Step 5: Solve for the final temperature

Now, we will isolate Tf and solve for the final temperature. (50 g * 4.18 J/g°C * Tf) - (50 g * 4.18 J/g°C * 25°C) = (7.25 g * 0.386 J/g°C * 100.1°C) - (7.25 g * 0.386 J/g°C * Tf) On solving the above equation, we get: Tf ≈ 28.3°C The final shared temperature of the copper and water in the insulated container is approximately 28.3°C.

Key Concepts

Specific Heat Capacity

When studying heat transfer, understanding the concept of specific heat capacity is crucial. Specific heat capacity refers to the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius. It's like how much effort it takes to warm up a substance by each degree.

The specific heat capacity varies from material to material:

• Water, with a specific heat capacity of 4.18 J/g°C, takes a lot of energy to change its temperature. This is why water is commonly used in home heating systems and as a coolant.
• Copper, on the other hand, has a much lower specific heat capacity of 0.386 J/g°C. This means copper heats up and cools down more quickly than water.

In our exercise, these properties determine how the final temperature is calculated when the two substances interact. Understanding these differences is fundamental in predicting how systems reach equilibrium.

Thermal Equilibrium

Thermal equilibrium is the state reached when two substances in contact no longer exchange heat, resulting in a uniform temperature throughout the system.

This occurs naturally when a hot object and a cool object touch. The hot object loses heat, while the cool object absorbs it until they both reach the same temperature.

In the exercise:

• The copper initially at 100.1°C and the water at 25.0°C exchange heat.
• This exchange results in a final uniform temperature of around 28.3°C.

Reaching thermal equilibrium means no more net heat flow occurs between the copper and water. This concept is key in designing sustainable systems, like thermal energy storage, where maintaining equilibrium is necessary for efficiency.

Calorimetry

Calorimetry is the science of measuring heat transfer in physical and chemical processes. It helps us understand how much heat is absorbed or released during these processes and is a vital tool in thermodynamics.

In a calorimetry problem like this exercise, we aim to find the final temperature after two substances have mixed:

• This involves calculating the heat lost by the hot object (copper) and the heat gained by the cool object (water).
• Setting these heat expressions equal helps find the final equilibrium temperature.

The formula used, derived from calorimetry principles, ensures the conservation of energy within the system. By understanding calorimetry, you can solve real-world heat transfer problems, such as optimizing the efficiency of a heater or determining how much heat a dessert absorbs when chilling.