Problem 42 - College Physics

Question

\(\cdot\) You have 750 \(\mathrm{g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

Add 1950 grams of boiling water.

Step 1: Understand the Concept

In this problem, we are dealing with heat exchange. The idea is that energy lost by the hot water while cooling down to the final temperature will be equal to the energy gained by the cold water as it heats up to the final temperature. We will use the principle of conservation of energy.

Step 2: Identify Key Equations

The key equation for heat gained or lost is \[ q = mc\Delta T \]where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity (for water it's 4.18 J/g°C), and \( \Delta T \) is the change in temperature.

Step 3: Set Up the Equation

Let \( m_h \) be the mass of the hot water to be added. The heat lost by the hot water is:\[ q_h = m_h \times c \times (100.0^{\circ} - 75.0^{\circ}) \]The heat gained by the cold water is:\[ q_c = 750 \times c \times (75.0^{\circ} - 10.0^{\circ}) \]Set \( q_h = q_c \) for conservation of energy.

Step 4: Simplify and Solve the Equation

Equating the two expressions gives:\[ m_h \times c \times 25 = 750 \times c \times 65 \]The specific heat \( c \) cancels out:\[ m_h \times 25 = 750 \times 65 \]Thus, solve for \( m_h \):\[ m_h = \frac{750 \times 65}{25} \]

Step 5: Calculate the Mass of Boiling Water Needed

Perform the calculation:\[ m_h = \frac{48750}{25} = 1950 \]Therefore, 1950 grams of boiling water must be added to achieve the desired final temperature.

Key Concepts

Conservation of Energy

The principle of conservation of energy is a fundamental concept in physics that tells us that energy cannot be created or destroyed. Instead, it can only change forms. In the context of our problem, this means that the heat energy lost by the hot water is equal to the heat energy gained by the cold water. This is crucial to find out how much boiling water we need to add.

By setting the heat lost by the hot water equal to the heat gained by the cold water, we can solve for the unknown mass of the hot water. This ensures that energy is balanced in the closed system of the insulated beaker, where no heat is lost to the surroundings.

Specific Heat Capacity

Specific heat capacity is a property of a substance that tells us how much heat energy is needed to raise the temperature of one gram of the substance by one degree Celsius.

For water, this value is 4.18 J/g°C. This means it takes 4.18 joules of energy to raise one gram of water by one degree Celsius. Knowing this value allows us to calculate the amount of heat energy exchanged during the process of heating or cooling.

It’s essential to understand this concept because it explains how different materials respond to heat. In our exercise, using the specific heat capacity of water, we plug it into the heat equation to calculate the energy changes when mixing different temperatures of water.

Temperature Change

Temperature change is represented as \( \Delta T \) in the heat equation \( q = mc\Delta T \). It indicates how much the temperature of a substance has increased or decreased.

In our exercise, the hot boiling water undergoes a decrease in temperature from \( 100^{\circ} \text{C} \) to the final temperature \( 75^{\circ} \text{C} \). Meanwhile, the cold water's temperature increases from \( 10^{\circ} \text{C} \) to \( 75^{\circ} \text{C} \).

• The temperature change for the hot water: \( \Delta T = 100 - 75 \)
• For the cold water: \( \Delta T = 75 - 10 \)

Understanding this helps us apply changes in temperature to determine the heat energy transferred in each scenario. It’s a key step in the process of calculating the mass of boiling water needed.

Mass Calculation

Mass calculation can be easily done when we understand all the above concepts and have our equations correctly set up. In our exercise, we calculated the mass of the boiling water needed to reach a desired final temperature with the equation:

• \[ m_h \times 25 = 750 \times 65 \]
• Solving gives \( m_h = \frac{750 \times 65}{25} \)
• Therefore, \( m_h = 1950 \) grams

This calculation was straightforward because the specific heat capacity \( c \) was the same for both masses of water and cancels out, leaving a simple multiplication equation to solve for \( m_h \). Understanding this step shows how knowing a few key details about the properties of substances allows us to deduce necessary quantities in heat exchange processes.