Question

One beaker contains \(156 \mathrm{g}\) 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 water temperature is approximately \(47.8^{\circ} \text{C}\).

Step-by-step solution

Understand Heat Exchange

When two bodies at different temperatures are mixed, heat will flow from the hotter to the cooler body until thermal equilibrium is reached. The energy gained by the cooler body equals the energy lost by the hotter body, assuming no heat loss to the surroundings.

Identify the Variables

Let's define the variables: - Let the mass of the cooler water be \(m_1 = 156 \text{ g} \), and its initial temperature \(T_1 = 22^{\circ} \text{C} \).- Let the mass of the hotter water be \(m_2 = 85.2 \text{ g} \), and its initial temperature \(T_2 = 95^{\circ} \text{C} \).- Let the final temperature of the mixture be \(T_f \). - The specific heat capacity of water is \(c = 4.18 \text{ J/g}\cdot\text{°C} \centerdot \).

Set Up Heat Exchange Equation

The heat lost by the hot water \((m_2c(T_2 - T_f))\) will be equal to the heat gained by the cold water \((m_1c(T_f - T_1))\). Thus, the equation is:\[ m_2c(T_2 - T_f) = m_1c(T_f - T_1) \]

Cancel Out Specific Heat Capacity

Since the specific heat capacities are the same, they can be canceled out from the equation, simplifying it to:\[ m_2(T_2 - T_f) = m_1(T_f - T_1) \]

Solve for Final Temperature

Substitute the values into the equation:\[ 85.2(95 - T_f) = 156(T_f - 22) \]Distribute and solve:\[ 8094 - 85.2T_f = 156T_f - 3432 \]Combine like terms and solve for \(T_f\):\[ 8094 + 3432 = 156T_f + 85.2T_f \]\[ 11526 = 241.2T_f \]\[ T_f = \frac{11526}{241.2} \approx 47.8^{\circ} \text{C} \]

Verify Solution

Ensure that both masses and temperatures satisfy the heat exchange equation, confirming the accuracy of our calculations.

Key concepts

Thermal Equilibrium

When we talk about mixing two bodies of water at different temperatures, the concept of thermal equilibrium takes center stage. Essentially, it implies that heat will naturally transfer from the warmer water to the cooler water. This process will continue until both bodies reach the same temperature, which is when they achieve thermal equilibrium.

It’s important to remember that the amount of heat lost by the warmer water is precisely equal to the amount of heat gained by the cooler water. This balance is a crucial principle in physics and is referred to as the law of conservation of energy. However, in practical scenarios, we assume that no heat is lost to the surrounding environment for simple calculations.
When you mix two different samples of liquid, say two separate volumes of water both at different temperatures, thermal equilibrium helps us predict the outcome. This result is the final temperature, leading us to the next core concept, specific heat capacity.

Specific Heat Capacity

Specific heat capacity is a significant factor in temperature changes during heat exchange between substances. It essentially tells us how much energy is required to change the temperature of 1 gram of a substance by 1 degree Celsius. For water, this value is commonly known to be 4.18 Joules/gram°C.

This constant is significant in calculating how much heat energy is involved when the heat is exchanged between the two bodies of water. It emphasizes the efficiency with which a substance can absorb heat. When dealing with water, we find it to be quite efficient at storing heat due to its relatively high specific heat capacity compared to many other materials.
In our exercise of mixing water at two different temperatures, the specific heat capacity plays a role in making sure our final calculations accurately reflect how water behaves. Knowing this property of water, we can set up the necessary equations to predict the final equilibrium temperature, something easily calculated once we understand how to perform temperature calculations.

Temperature Calculation

The final piece of the puzzle in understanding our original exercise scenario is calculating the eventual mixed temperature, or temperature calculation. This involves using the principle set by thermal equilibrium and specific heat capacity.

In simplified terms, the equation used to find the final temperature of the mixed water is based on setting the heat lost by the warmer water equal to the heat gained by the cooler water. The equation: \[ m_2c(T_2 - T_f) = m_1c(T_f - T_1) \] will allow us to solve for the unknown final temperature \( T_f \). In this equation, we can cancel out the specific heat capacity \( c \) (because it is constant for both masses of water), making it a bit easier to solve.
Substituting the known values in, we can rearrange and solve for \( T_f \). This problem-solving method allows us not only to understand how mixing works but also gives us insight into the heat dynamics involved, helping us grasp the concept more intuitively.