Question

Equal masses of liquid A, initially at \(100^{\circ} \mathrm{C},\) and liquid \(\mathrm{B}\), initially at \(50^{\circ} \mathrm{C},\) are combined in an insulated container. The final temperature of the mixture is \(80^{\circ} \mathrm{C}\). All the heat flow occurs between the two liquids. The two liquids do not react with each other. Is the specific heat of liquid A larger than, equal to, or smaller than the specific heat of liquid \(\mathrm{B}\) ?

Short answer

Answer: Liquid A has a higher specific heat capacity than Liquid B.

Step-by-step solution

Define the variables and equations

Let's denote the specific heat capacities of liquid A and liquid B as \(c_A\) and \(c_B\). The masses of both liquids are equal, so let's denote their masses as \(m\). The initial and final temperatures are given as: - Liquid A: \(T_{A_i} = 100^{\circ} \mathrm{C}\) - Liquid B: \(T_{B_i} = 50^{\circ} \mathrm{C}\) - Final temperature: \(T_f = 80^{\circ} \mathrm{C}\) Since the heat lost by liquid A is equal to the heat gained by liquid B, we can write: \(Q_{A} = Q_{B}\) Using the equation for heat flow, \(Q = mc\Delta T\), we can write: \(m c_A (T_{A_i} - T_f) = m c_B (T_f - T_{B_i})\)

Cancel out the masses

Since the masses of both liquids are equal, we can cancel them out from the equation: \(c_A (T_{A_i} - T_f) = c_B (T_f - T_{B_i})\)

Plug in the given temperatures and solve

Now, let's plug in the values of \(T_{A_i}, T_{B_i}\) and \(T_f\) into the equation: \(c_A (100 - 80) = c_B (80 - 50)\) Simplifying the equation: \(20c_A = 30c_B\) Dividing both sides of the equation by 10, we get: \(2c_A = 3c_B\)

Determine the specific heat relationship

Now, we can see that: \(c_B = \frac{2}{3} c_A\) Since \(\frac{2}{3} < 1\), we can conclude that the specific heat capacity of liquid A is larger than the specific heat capacity of liquid B.

Key concepts

Thermal Equilibrium

Understanding thermal equilibrium is key to decoding why materials reach a certain temperature when mixed. Thermal equilibrium refers to the condition when two materials or objects in contact with each other exchange heat until they reach the same temperature. At this point, there is no net heat flow between them; heat transfer has ceased, and their temperatures remain constant assuming there is no heat loss to the surroundings.

In our exercise, when liquid A at a higher initial temperature and liquid B at a lower initial temperature are combined in an insulated container, heat flows from A to B. Eventually, they both settle at a final temperature of 80°C. This final temperature is a result of the system reaching thermal equilibrium. The concept solidifies the understanding that, in an isolated system, temperatures will homogenize over time due to heat flow.

Heat Flow

Heat flow, often measured in joules in the metric system, is the transfer of thermal energy from a body of higher temperature to one of lower temperature. It's a fundamental concept in thermodynamics and is vital in our specific heat capacity question. To determine the direction and magnitude of heat flow, we use the formula \(Q = mc\Delta T\), where \(Q\) is the heat flow, \(m\) is the mass of the substance, \(c\) represents specific heat capacity, and \(\Delta T\) denotes the temperature change.

When we equalized the heat lost by liquid A to the heat gained by liquid B in our solution, we applied this formula. The formula assumes no heat lost to the environment, which is why our mixture was placed in an insulated container. This allowed us to directly compare the heat capacities of the two substances by their temperature changes when they reached thermal equilibrium.

Temperature Change

Temperature change is a measure of the difference in temperature as a substance absorbs or releases heat. In the context of our exercise, temperature change is crucial for calculating heat flow and, subsequently, comparing the specific heat capacities of different substances. It's calculated using the formula \(\Delta T = T_{final} - T_{initial}\).

As we saw in our step-by-step solution, liquid A cooled from 100°C to the final equilibrium temperature of 80°C, a change of 20°C, while liquid B warmed up from 50°C to 80°C, a change of 30°C. These temperature changes, when plugged into the heat flow equation in conjunction with the respective specific heat capacities, allowed us to discern the relationship between the specific heats of liquid A and B, ultimately leading us to conclude that liquid A has a larger specific heat capacity than liquid B.