Question

A \(30.0\) -g sample of water at \(280 . \mathrm{K}\) is mixed with \(50.0 \mathrm{~g}\) water at \(330 . \mathrm{K}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Short answer

The final temperature of the mixture, assuming no heat loss to the surroundings, is \(405 K\).

Step-by-step solution

State the known values and equation to calculate heat transfer

We know the masses and initial temperatures of the two water samples, and we'll use the equation q = mcΔT for heat transfer calculations. The specific heat capacity of water, c, is 4.18 J/(g·K). Sample 1: m1 = 30 g, T1_initial = 280 K Sample 2: m2 = 50 g, T2_initial = 330 K

Find the heat gained/lost by the samples based on their final temperature

We need to find the final temperature (T_final) of the mixture and write the equation for both samples considering that the heat gained is equal to the heat lost: q1 = -q2 For Sample 1, we have: q1 = m1 * c * (T_final - T1_initial) And for Sample 2, we have: q2 = m2 * c * (T2_initial - T_final) Since q1 = -q2, we can now write the equation as: m1 * c * (T_final - T1_initial) = -m2 * c * (T2_initial - T_final)

Solve for the final temperature

Now, we will solve the equation for T_final. Notice that c is present in both terms, so we can divide the entire equation by c: m1 * (T_final - T1_initial) = -m2 * (T2_initial - T_final) Now, let's simplify the equation and solve for T_final: 30 * (T_final - 280) = -50 * (330 - T_final) Expanding the equation and combining the terms, we get: 30 * T_final - 8400 = -50 * 330 + 50 * T_final 30 * T_final - 8400 = -16500 + 50 * T_final Now we can isolate T_final: 20 * T_final = 8100 Finally, we can find the T_final: T_final = 8100 / 20 = 405 K

State the answer

The final temperature of the mixture, assuming no heat loss to the surroundings, is 405 K.

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics involving the movement of thermal energy from one object or substance to another. In the context of our exercise, heat transfer plays a key role because when two bodies of water at different temperatures are mixed, heat will flow from the warmer body to the cooler body until thermal equilibrium is reached.

From a microscopic perspective, this process happens as the faster-moving molecules of the warmer water collide with the slower-moving molecules of the cooler water, transferring kinetic energy in the process. This results in the warmer water cooling down while the cooler water heats up. In the equation provided in the step-by-step solution, the term q represents the quantity of heat gained or lost by each water sample during mixing which is crucial for calculating the final temperature.

Specific Heat Capacity

Specific heat capacity, often symbolized as c, is a property that indicates how much heat energy is required to raise the temperature of a unit mass of a substance by one Kelvin (or one degree Celsius). A high specific heat capacity means the substance can absorb a lot of heat without experiencing a significant change in temperature.

Water has a high specific heat capacity of approximately 4.18 J/(g·K), which means it takes 4.18 joules of energy to raise the temperature of one gram of water by one Kelvin. This property is key to the exercise because it allows us to calculate the heat exchange between the different masses of water using the formula q = mcΔT. The consistent value of c across both water samples simplifies our calculations as it cancels out when we equate the heat lost by the hot water to the heat gained by the cool water (q1 = -q2).

Thermal Equilibrium

Thermal equilibrium is the state reached when two objects in contact no longer exchange heat, meaning they have the same temperature. It's based on the zeroth law of thermodynamics which states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.

In our exercise, when the 30.0-g sample of water at 280 K is mixed with 50.0 g water at 330 K, the two samples will exchange heat until they reach the same final temperature, achieving thermal equilibrium. The steps provided derive this final temperature mathematically by equating the heat lost by the warmer sample with the heat gained by the cooler sample, resulting in a final temperature where no more heat exchange occurs. The calculated final temperature represents the point of thermal equilibrium for the mixture.