Question

One beaker contains 156 g of water at \(22^{\circ} \mathrm{C},\) and a second beaker contains \(85.2 \mathrm{g}\) of water at \(95^{\circ} \mathrm{C}\) The water in the two beakers is mixed. What is the final water temperature?

Short answer

The final temperature is approximately \(47.8^{\circ} \text{C}\).

Step-by-step solution

Understand the Problem

We are tasked with finding the final temperature when two different masses of water at different temperatures are mixed. This is a problem involving the principle of conservation of energy, specifically thermal energy.

Identify Known Values

We have the following data: - Mass of the first beaker, \(m_1 = 156 \, \text{g}\)- Temperature of the first beaker, \(T_1 = 22^{\circ} \text{C}\)- Mass of the second beaker, \(m_2 = 85.2 \, \text{g}\)- Temperature of the second beaker, \(T_2 = 95^{\circ} \text{C}\)

Apply the Principle of Conservation of Energy

When the two samples are mixed, the total energy must be conserved, meaning the heat lost by the warmer water must equal the heat gained by the cooler water.Use the formula:\[m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0\]where:- \(c\) is the specific heat capacity of water (\(4.18 \, \text{J/g°C}\))- \(T_f\) is the final temperature.

Simplify the Equation

Since the specific heat capacity \(c\) and the unit of mass are the same for both, they cancel each other out, simplifying our equation to:\[m_1 \cdot (T_f - T_1) + m_2 \cdot (T_f - T_2) = 0\]

Solve for Final Temperature \(T_f\)

Rearrange the equation to:\[m_1 \cdot T_f - m_1 \cdot T_1 + m_2 \cdot T_f - m_2 \cdot T_2 = 0\]Combine terms with \(T_f\):\[(m_1 + m_2) \cdot T_f = m_1 \cdot T_1 + m_2 \cdot T_2\]Solve for \(T_f\):\[T_f = \frac{m_1 \cdot T_1 + m_2 \cdot T_2}{m_1 + m_2}\]

Substitute Known Values into Equation

Substitute the known values into our equation:\[T_f = \frac{156 \cdot 22 + 85.2 \cdot 95}{156 + 85.2}\]Calculate the numerator: \(156 \times 22 + 85.2 \times 95 = 3432 + 8094 = 11526.2\)Calculate the denominator: \(156 + 85.2 = 241.2\)Therefore:\[T_f = \frac{11526.2}{241.2} \approx 47.8^{\circ} \text{C}\]

Conclusion

The final temperature of the mixture is approximately \(47.8^{\circ} \text{C}\) after mixing the two different water samples at different initial temperatures.

Key concepts

Conservation of Energy

When mixing two liquids at different temperatures, the concept of **Conservation of Energy** is an essential principle. Specifically, the focus is on the conservation of thermal energy. This principle states that the total energy in a closed system remains constant.
In our scenario, when two beakers of water are mixed, the warmer water will lose energy while the cooler water gains energy. According to conservation of energy, the amount of heat lost by the hot water equals the amount of heat gained by the cold water.
This ensures no energy is created or destroyed, and instead, it is simply transferred from one body to another. This transfer continues until thermal equilibrium is reached, meaning both masses of water reach the same temperature. The equation for conservation of thermal energy is:

• \[m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0\]
• Where:
  • \(m_1\) and \(m_2\) are masses of the water.
  • \(T_1\) and \(T_2\) are initial temperatures.
  • \(c\) is the specific heat capacity of water.
  • \(T_f\) is the final temperature we want to find.

By understanding this principle, we can solve temperature problems involving energy exchange between substances.

Specific Heat Capacity

**Specific Heat Capacity** is a critical concept in understanding how different substances absorb heat. It refers to the amount of heat energy required to change the temperature of a unit mass of a substance by 1 degree Celsius (\(^\circ\)C). In simpler terms, it measures a material's heat absorption capacity.
In our exercise, the specific heat capacity of water is used, which is approximately 4.18 J/g°C. This value tells us how much heat energy is needed to change the temperature of 1 gram of water by 1°C.
When calculating the final temperature of mixed liquids, knowing the specific heat capacity helps in understanding how each liquid absorbs or loses heat.
Because liquid water has a high specific heat capacity, it is very effective at absorbing heat with little change in temperature. This makes water an excellent substance for thermal equilibrating processes. In the equation used for finding the final temperature, specific heat capacity ensures accuracy in understanding temperature changes and energy transfer.
It is important to realize that for problems like this, if all substances were of the same material (like water in this case), the value of specific heat capacity cancels out in calculations, simplifying the process.

Mixing of Liquids

When dealing with the **Mixing of Liquids**, it is essential to understand how different initial conditions will affect the final equilibrium state. The process is straightforward once you know the mass and temperature of each liquid.
The goal is to find a final temperature that reflects the thermal equilibrium after mixing. In our example, two beakers of water with different temperatures are combined. Relevant factors in determining the final temperature include:

• The mass of the liquids being mixed. Greater mass has more influence on the final temperature.
• The initial temperatures. The greater the temperature difference, the more heat exchange will occur.

The process begins with the warmer liquid transferring some of its heat to the cooler one, resulting in an intermediate temperature. By employing the conservation of energy and recognizing that water's properties allow us to simplify calculations, we derive the equation:

• \[T_f = \frac{m_1 \cdot T_1 + m_2 \cdot T_2}{m_1 + m_2}\]

Through this, we calculate the common temperature both masses of water will reach once they have exchanged heat thoroughly. Such insights are vital for practical applications like adjusting climate conditions in larger systems or understanding natural processes such as ocean currents.