Calculating Equilibrium Temperature When Mixing Water

This problem involves the mixing of two quantities of water at different temperatures. When substances at different temperatures are mixed in an insulated system (where no heat is lost or gained from the surroundings), heat energy transfers from the hotter substance to the colder substance until they reach a common final equilibrium temperature. The fundamental principle governing this process is that the heat lost by the hot substance equals the heat gained by the cold substance.

Understanding the Principle of Heat Exchange

The amount of heat energy (\(Q\)) transferred is given by the formula:

\(Q = mc\Delta T\)

Where:

• \(m\) is the mass of the substance.
• \(c\) is the specific heat capacity of the substance.
• \(\Delta T\) is the change in temperature (\(T_{final} - T_{initial}\)). For heat loss, \(\Delta T\) is often written as \(T_{initial} - T_{final}\).

In this scenario, we are mixing hot water and cold water. Assuming the specific heat capacity (\(c\)) of water is constant over the given temperature range and the density (\(\rho\)) of water is also constant (approximately 1 kg/L), we can determine the mass of each quantity of water from its volume.

Given Information:

• Volume of hot water (\(V_1\)) = 2 litres
• Initial temperature of hot water (\(T_1\)) = 190°C
• Volume of cold water (\(V_2\)) = 6 litres
• Initial temperature of cold water (\(T_2\)) = 20°C
• No heat is lost to the surroundings (adiabatic mixing).

Assuming density of water \(\rho \approx 1 \text{ kg/L}\):

• Mass of hot water (\(m_1\)) \(= V_1 \times \rho = 2 \text{ L} \times 1 \text{ kg/L} = 2 \text{ kg}\)
• Mass of cold water (\(m_2\)) \(= V_2 \times \rho = 6 \text{ L} \times 1 \text{ kg/L} = 6 \text{ kg}\)

Setting up the Heat Balance Equation

According to the principle of heat exchange in an isolated system:

\(\text{Heat lost by hot water} = \text{Heat gained by cold water}\)

\(m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)\)

Where \(T_f\) is the final equilibrium temperature we need to find.

Since the specific heat capacity \((c)\) is the same for both (water), we can cancel it from both sides of the equation:

\(m_1 (T_1 - T_f) = m_2 (T_f - T_2)\)

Solving for the Final Temperature

Now, we substitute the known values into the equation:

\(2 \text{ kg} \times (190^\circ\text{C} - T_f) = 6 \text{ kg} \times (T_f - 20^\circ\text{C})\)

Expand both sides of the equation:

\(2 \times 190 - 2 \times T_f = 6 \times T_f - 6 \times 20\)

\(380 - 2T_f = 6T_f - 120\)

Now, rearrange the terms to group \(T_f\) on one side and constants on the other:

\(380 + 120 = 6T_f + 2T_f\)

\(500 = 8T_f\)

Finally, solve for \(T_f\):

\(T_f = \frac{500}{8}\)

\(T_f = 62.5^\circ\text{C}\)

Thus, the final equilibrium temperature after mixing the superheated water at 190°C and the cold water at 20°C is 62.5°C.

Summary of Calculation Steps

Revision Table: Key Concepts in Mixing

Additional Information: Assumptions and Considerations

Several assumptions were made in solving this heat mixing problem:

• Constant Specific Heat Capacity: We assumed that the specific heat capacity of water remains constant over the temperature range from 20°C to 190°C. In reality, it varies slightly with temperature, but this assumption is common for simplified calculations.
• Constant Density: We assumed the density of water is constant at 1 kg/L. Density also varies slightly with temperature and pressure, but this is a reasonable approximation for this problem.
• No Phase Change: The term "superheated water at 190°C" indicates liquid water at a pressure above its saturation pressure at 190°C. The problem implicitly assumes that no boiling or phase change occurs during or after mixing, which would involve latent heat transfer and a more complex calculation. The final temperature (62.5°C) is well below the boiling point at standard pressure, confirming no boiling in the final state.
• Perfect Mixing: It is assumed that the two volumes of water mix completely and instantly to reach a uniform final temperature.
• Isolated System: The problem states "no heat is lost," meaning the system (the mixed water) is isolated from the surroundings, preventing any heat transfer to or from anything else.