Question

Find the net change in entropy when \(100 . \mathrm{g}\) of water at \(0^{\circ} \mathrm{C}\) is added to \(100 . \mathrm{g}\) of water at \(100 .{ }^{\circ} \mathrm{C}\)

Short answer

Answer: The net change in entropy when 100g of water at 0°C is added to 100g of water at 100°C is 4180 J/K.

Step-by-step solution

Convert temperatures to Kelvin

To avoid inconsistencies, we should convert the initial temperatures from Celsius to Kelvin. To do this, we add 273.15 to each Celsius value: $$ T_{1} = 0^{\circ}C + 273.15 = 273.15 \ K \\ T_{2} = 100^{\circ}C + 273.15 = 373.15 \ K $$

Calculate the Final Temperature

Before we calculate the change in entropy, we need to find the final temperature after mixing two masses of water. Since there is no heat loss to the surroundings, the sum of the heat absorbed and released will be zero: $$ \Delta Q_{1} + \Delta Q_{2} = 0 $$ We can express the heat transfer as: $$ mc\Delta T_{1} = -mc\Delta T_{2} $$ Where \(\Delta T_{1}\) and \(\Delta T_{2}\) are the temperature changes of the water masses. Since the masses and specific heat are equal, we can simplify and solve for the final temperature, \(T_{f}\): $$ \Delta T_{1} = -\Delta T_{2} \\ T_{f}-T_{1} = - (T_{f}-T_{2}) \\ 2T_{f} = T_{1} + T_{2} \\ T_{f} = \frac{T_{1} + T_{2}}{2} = \frac{273.15 \ K + 373.15 \ K}{2} = 323.15 \ K $$

Calculate the Change in Entropy

Now, we can calculate the change in entropy for each water mass using the formula mentioned earlier: $$ \Delta S = mc\Delta T $$ For mass 1: $$ \Delta S_{1} = (100 \ g)(4.18 \ J/g \cdot K)(323.15 \ K - 273.15 \ K) $$ For mass 2: $$ \Delta S_{2} = (100 \ g)(4.18 \ J/g \cdot K)(373.15 \ K - 323.15 \ K) $$

Calculate the Net Change in Entropy

Finally, we can sum ∆S1 and ∆S2 to find the net change in entropy: $$ \Delta S_{net} = \Delta S_{1} + \Delta S_{2} $$ Evaluate for ∆S1 and ∆S2 we found in step 3: $$ \Delta S_{net} = (100 \times 4.18 \times (323.15-273.15)) + (100 \times 4.18 \times (373.15-323.15)) = 2090 + 2090 = 4180 \ J/K $$ The net change in entropy when 100g of water at 0°C is added to 100g of water at 100°C is 4180 J/K.

Key concepts

Thermodynamics

Understanding thermodynamics is essential to grasp how energy is transformed in various systems, which is the backbone of the entropy change problem.
At the heart of thermodynamics lie several key concepts and laws. The First Law of Thermodynamics, for instance, which is essentially a statement of energy conservation, tells us that energy cannot be created or destroyed, only transformed from one form to another. One common process we see this law in action is during the heating or cooling of substances - where thermal energy is converted into internal energy resulting in temperature changes.

Meanwhile, the Second Law of Thermodynamics introduces entropy, a measure of the disorder or randomness in a system. It states that the total entropy of an isolated system can never decrease over time. In our exercise, we witness this law as we calculate the change in entropy when two bodies of water at different temperatures mix. Such a process naturally progresses towards equilibrium, creating a net increase in the system's entropy, which aligns with the law's prediction of directional change towards disorder.

Heat Transfer

In our problem, the core phenomenon at play is heat transfer, which occurs when two bodies at different temperatures come into contact and exchange heat, striving to reach a thermal equilibrium.
There are three modes of heat transfer: conduction, convection, and radiation. Our example primarily revolves around conduction, the direct transfer of heat through a substance when there is a difference in temperature. This is how heat energy moves from the hotter water to the cooler water until they both reach the same temperature.

Understanding the mechanism of heat transfer helps in grasping why the final temperature of the mixed water system is an average of the initial temperatures. It's also foundational in calculating the net change in entropy for the system as heat transfer directly influences the change in entropy within the water.

Specific Heat

Specific heat is a substance's capacity to absorb heat for a given increase in temperature. It's the amount of heat per unit mass required to raise the temperature by one Kelvin.
The formula to calculate the amount of heat absorbed or released by a substance as its temperature changes is \( Q = mc\Delta T \), where \( Q \) is the heat transfer, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.

In the exercise, we deal with water, which has a high specific heat. This characteristic allows water to absorb a lot of heat with little change in temperature. It's why we can use the same value for specific heat for both masses of water when calculating the change in entropy. If we were working with substances with different specific heats, the calculations would become more complex.

Temperature Conversion

The temperature conversion from Celsius to Kelvin is a crucial step in thermodynamic calculations and needs to be accurate to avoid errors. Kelvin, being the SI unit for temperature, is preferred in scientific equations because it provides an absolute scale where zero is the absolute zero, the point where no thermal energy can be extracted from a system.
To convert Celsius to Kelvin, you add 273.15 to the Celsius value. This seems straightforward, but it's essential to accurate entropy calculations, as seen in our problem.

Moreover, understanding temperature scales and conversions can also help prevent conceptual mistakes, such as confusing temperature difference with a specific temperature value or overlooking the absolute nature of temperature measurements in thermodynamic laws.