Question

Two liquids of equal volume are thoroughly mixed. If their specific heat are \(c_{1}, c_{2}\), temperatures \(\theta_{1}, \theta_{2}\) and densities \(\mathrm{d}_{1}, \mathrm{~d}_{2}\) respectively. What is the final temperature of the mixture ? (A) \(\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right)\right\\}\) (B) \(\left\\{\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}\) (C) \(\left\\{\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right) /\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right)\right\\}\) (D) \(\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}\)

Short answer

Based on the provided step-by-step solution, the correct short answer to the question is: The final temperature of the mixture is given by: \( \left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\} \) (Option D)

Step-by-step solution

Determine the heat gained and lost by each liquid

First, let's determine the heat gained by liquid 1 (Q1) and the heat lost by liquid 2 (Q2). We'll assume a positive Q represents heat gained and a negative Q represents heat lost. Since the liquids have equal volume, we can say that: Q1 = mass1 × specific heat1 × (Tf - T1) Q2 = mass2 × specific heat UIScreen2FFinHLOJQU-BB]]></

Key concepts

Specific Heat Capacity

Specific heat capacity is a property of a substance that tells us how much heat energy is required to raise the temperature of a unit mass of that substance by one degree Celsius. It is often denoted by the symbol \( c \). The specific heat capacity depends on the nature of the substance, for example, water has a specific heat capacity of approximately \( 4.18 \, \text{J/g°C} \), which means it takes 4.18 joules to raise the temperature of one gram of water by one degree Celsius.

• A higher specific heat capacity means the substance can absorb more heat without a significant change in temperature.
• In practical scenarios, materials with high specific heat capacities are often used in thermal storage applications.

In the context of mixing two liquids, specific heat capacities \( c_1 \) and \( c_2 \) help determine how much energy each liquid can absorb or release, affecting the final temperature of the mixture.

Density and Volume

Density is a measure of how much mass is contained in a given volume. It is usually expressed in units such as \( \text{kg/m}^3 \). The formula for density \( \rho \) is given by \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume.

• Equal volumes of different substances will have different masses if their densities are different.
• In mixing problems, knowing the density helps determine the mass from the given volume.
• Understanding densities is crucial in calculating the heat transfer in mixed liquids.

For two liquids of equal volume, with densities \( d_1 \) and \( d_2 \), different masses will still apply even if the volumes are the same. When solving the thermal equilibrium problem, it is essential to consider this mass difference due to density to appropriately calculate heat transfer.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. In thermal equilibrium scenarios, such as the mixing of two liquids, heat will flow from the hotter liquid to the cooler one until both reach the same temperature.

• The formula for heat transfer for a substance is given by \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.
• When two objects are in thermal contact, they will exchange heat until they reach a uniform temperature, known as thermal equilibrium.

In liquid mixing problems, the heat lost by the warmer liquid equals the heat gained by the cooler liquid due to the law of conservation of energy. Thus, in the given problem, the energy balance equation combines the heat capacities, densities, and initial temperatures to find the final temperature of the liquid mixture.