Problem 42 - College Physics

Question

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

1950 g of boiling water must be added.

Step 1: Understanding the Problem

We are given two masses of water at different temperatures and are required to determine how much boiling water must be added to a known quantity of water at a lower temperature to achieve a specified final temperature. The mass of boiling water is our unknown variable.

Step 2: Using the Concept of Heat Transfer

The concept of heat transfer states that the heat lost by the hot water will equal the heat gained by the cold water. Mathematically, this is represented as:\[ m_c \cdot c \cdot \Delta T_c = m_h \cdot c \cdot \Delta T_h \]where \(m_c\) and \(m_h\) are the masses of cold and hot water, \(c\) is the specific heat (which cancels out in both sides as it is the same), and \(\Delta T_c\) and \(\Delta T_h\) are the temperature changes of the cold and hot water respectively.

Step 3: Setting Up the Equation

Let's denote the mass of the boiling water to be added as \(m_h\). The mass of the cold water is \(750 \mathrm{~g}\). The initial and final temperatures of the cold water are \(10.0^{\circ} \mathrm{C}\) and \(75^{\circ} \mathrm{C}\), a change of \(65^{\circ} \mathrm{C}\), and for the hot water, the initial and final temperatures are \(100.0^{\circ} \mathrm{C}\) and \(75^{\circ} \mathrm{C}\), a change of \(-25^{\circ} \mathrm{C}\). Thus, the equation becomes:\[ 750 \cdot (75 - 10) = m_h \cdot (75 - 100) \]

Step 4: Solving the Equation

Simplify and solve the equation:\[ 750 \cdot 65 = m_h \cdot (-25) \]\[ 48750 = -25m_h \]\[ m_h = \frac{48750}{25} = 1950 \mathrm{~g} \]

Step 5: Verifying the Solution

To ensure the solution is correct, calculate the total energy absorbed and released. The cold water gains:\[ 750 \cdot 65 = 48750 \text{ energy units} \]The hot water loses:\[ 1950 \cdot 25 = 48750 \text{ energy units} \]Both values are equal, confirming the solution's validity.

Key Concepts

Thermal Equilibrium

Thermal equilibrium occurs when two or more objects reach a state where they have the same temperature. In the context of mixing water, it means that the boiling water and the colder water eventually mix to form a uniform temperature, without any further exchange of heat energy. This state is crucial because it dictates the final temperature of the mixed water. When two bodies at different temperatures come into contact, the heat energy from the hotter body will transfer to the cooler one until they reach the same temperature. This is the basis for calculating how much boiling water is needed to achieve a desired temperature when mixed with colder water.
In our exercise, the hot boiling water at 100°C and the cold water at 10°C achieve thermal equilibrium at 75°C. Calculating this requires understanding how much heat each body will lose or gain until they match temperatures.

Specific Heat

Specific heat is a property of a substance that indicates the amount of heat required to change the temperature of 1 gram of the substance by 1°C. For water, this value is consistently high at about 4.18 joules per gram per degree Celsius. Specific heat is central to understanding how materials react to temperature changes.
In the given problem, we deal with water, so we use its specific heat value in calculations. However, the specifics of this problem simplify the use of specific heat because it cancels out on both sides of the equation when equal masses and the same type of material are involved. This means, instead of doing detailed calculations for each body of water, we can focus on the balance of heat transfer without directly involving specific heat in our final computation.

Temperature Change

Temperature change (\(\Delta T\)) is the difference between the final and initial temperatures of a substance. It tells us how much a substance has been heated or cooled during a process. Understanding this concept is key for solving problems related to heat transfer.In the example exercise:

• The colder water sees a temperature change from 10°C to 75°C, which means an increase or rise in temperature (\(\Delta T_c = 75 - 10 = 65°C\)).
• The boiling water cools down from 100°C to 75°C, indicating a decrease or drop in temperature (\(\Delta T_h = 75 - 100 = -25°C\)).

Recognizing that one mass gains heat while the other loses it is fundamental, as it guides how we equate and solve these expressions mathematically.

Energy Conservation

The principle of energy conservation is a cornerstone of physics, stating that in an isolated system, energy cannot be created or destroyed, only transferred or transformed. This concept applies directly to thermal systems, where it predicts that the energy lost by hotter substances should equal the energy gained by cooler substances during heat exchange.
In the exercise, the balance of energy is expressed through the equation:\[ m_c \cdot c \cdot \Delta T_c = m_h \cdot c \cdot \Delta T_h \]Here, the energy gained by the cold water (\(750 \cdot 65 = 48750\)) equals the energy lost by the hot water (\(1950 \cdot 25 = 48750\)). This confirms the conservation of energy, ensuring that all heat energy from the hot water is transferred efficiently to the cold water, achieving the desired final temperature.