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Calculate the change in entropy of 250 \(\mathrm{g}\) of water heated slowly from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} .\) (Suggestion: Note that \(d Q=m c d T . )\)

Short Answer

Expert verified
The change in entropy (ΔS) is approximately 146.5 J/K.

Step by step solution

01

Understanding the Concept of Entropy Change

The entropy change (ΔS) for a substance when it is heated can be calculated using the formula ΔS = ∫(dQ/T), where dQ is the heat transfer to the substance and T is the absolute temperature. Since the process is slow (or quasi-static), it can be assumed the process is reversible, hence we can use this formula. Here, dQ can be expressed as mc dT, where m is the mass, c is the specific heat capacity, and dT is the change in temperature.
02

Converting Celsius to Kelvin

Convert the initial and final temperatures from Celsius to Kelvin by adding 273.15 (since K = °C + 273.15). The initial temperature (T_i) is 293.15 K (20.0°C + 273.15) and the final temperature (T_f) is 353.15 K (80.0°C + 273.15).
03

Calculating the Entropy Change

The entropy change can be calculated by integrating the expression for dQ/T = (mc dT)/T from T_i to T_f. Since c (specific heat capacity of water) is relatively constant over the range of temperatures, we integrate mc/T from T_i to T_f. Thus, the entropy change (ΔS) is mc ln(T_f/T_i)
04

Plugging in the Values

Insert the values into the entropy change formula (ΔS = m c ln(T_f/T_i)), with m = 0.250 kg (remember to convert grams to kilograms), c = 4184 J/(kg·K) (specific heat capacity of water), T_i = 293.15 K, and T_f = 353.15 K.
05

Execute the Calculation

Calculate ΔS using the previously obtained values. ΔS = 0.250 kg × 4184 J/(kg·K) × ln(353.15 K / 293.15 K).
06

Finding the Numerical Value

Now, compute the natural logarithm value and multiply by 0.250 kg and 4184 J/(kg·K) to find ΔS.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
At its core, thermodynamics is the science of energy, encompassing its transformation and the laws governing these processes. One of the fundamental concepts within thermodynamics is entropy, symbolized as 'S'. Entropy quantifies the degree of disorder or randomness in a system. The second law of thermodynamics states that in a closed system, the entropy will either increase or remain the same over time, signifying the direction of natural processes towards disorder.
In the context of our problem, heating water causes an increase in entropy. This is because as water molecules are heated, they move more rapidly, and their disorder increases. When calculating the entropy change of water as it is heated, we are examining how the energy transfer to the water (in the form of heat) affects its internal order, thus revealing the fundamental interplay between energy and disorder in thermodynamic processes.
Heat Transfer
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into various mechanisms, such as conduction, convection, and radiation. In our exercise, we deal with the heating of water, which involves the transfer of heat into the water, increasing its temperature.

When calculating entropy change due to heat transfer, it's essential to consider the specific manner in which heat is applied. In the given exercise, it's mentioned that the water is heated 'slowly', implying a quasi-static process, which is one where the system is nearly in equilibrium at all times. This slow heating ensures that the temperature of the water can be defined at any moment, simplifying the calculation of the entropy change during the process.
Specific Heat Capacity
The specific heat capacity (c) is a property of matter that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree in temperature. It's usually expressed in units of joules per kilogram per degree Kelvin (J/(kg·K)). In our calculation, we use water's specific heat capacity, which is relatively high, meaning that water can absorb a lot of heat before it increases in temperature.
In the step-by-step solution provided, the specific heat capacity is a crucial part of the formula for entropy change because it connects the amount of heat added to the temperature change experienced by the water. As the specific heat capacity is assumed to be constant over the small temperature range in the problem, this simplifies the integral needed to calculate the total entropy change.
Kelvin Temperature Scale
The Kelvin temperature scale is an absolute temperature scale starting at absolute zero, the point at which there is theoretically no thermal energy in a substance. Absolute zero is 0 K, which is equivalent to -273.15°C. Unlike the Celsius scale, where 0 is set at the freezing point of water, the Kelvin scale is essential in scientific calculations because it provides an absolute reference for thermal energy levels.

In thermodynamic formulas, such as entropy change, temperatures must be expressed in Kelvin to avoid negative values or division by zero. This is because such calculations often involve ratios or logarithms of temperatures, and the Kelvin scale ensures these operations are mathematically valid. Converting Celsius to Kelvin, as we have done in the exercise solution, is straightforward: add 273.15 to the Celsius temperature to get the equivalent Kelvin temperature.

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