Mix 4 kg of water at 80 °C and 6 kg of water at 20 °C.

The problem combines two portions of water of different temperatures: 4 kg at 80 °C and 6 kg at 20 °C. It is necessary to determine the change in entropy during the mixing process.

To solve this problem, we use the formula for entropy change: ΔS = Send - First,

where ΔS is the change in entropy, Skon is the entropy of the final state of the system, Snach is the entropy of the initial state of the system.

?ntropy can be calculated using the formula: S = Cpln(T) + Const,

where C is the heat capacity of the substance, T is the temperature in Kelvin, Const is a constant.

For each portion of water, we find its entropy:

• for 4 kg of water at 80 °C: S1 = 4 * 4184 * ln(80+273) + Const = 4 * 4184 * ln(353) + Const;
• for 6 kg of water at 20 °C: S2 = 6 * 4184 * ln(20+273) + Const = 6 * 4184 * ln(293) + Const.

When water is mixed, the temperature equalizes to an equilibrium state. In this case, the amount of heat transferred from a hotter portion to a colder one can be calculated using the formula: Q = m1 * C1 * (Tcon - Tav),

where Q is the amount of heat, m1 is the mass of a hotter portion of water, C1 is the heat capacity of water, Tkon is the final temperature of the equilibrium state, Tav is the average temperature of the initial portions of water.

The average temperature of the initial portions of water can be calculated using the formula: Tav = (m1 * T1 + m2 * T2) / (m1 + m2),

where m2 is the mass of a colder portion of water, T1 and T2 are the temperatures of the initial portions of water.

Thus, when mixing 4 kg of water at 80 °C and 6 kg of water at 20 °C we get:

• average temperature of the initial portions of water: Tav = (4 * 80 + 6 * 20) / (4 + 6) = 44 °C;
• the amount of heat transferred from a hotter portion to a colder one: Q = 4 * 4184 * (44 - 80) = -600448 J.

The change in entropy can be calculated as the difference between the entropy of the final and initial states: ΔS = Sfin - Initial = (S1 + S2) - Sinit = 4 * 4184 * ln(353) + 6 * 4184 * ln(293) + Const - (4 * 4184 * ln(80+273) + 6 * 4184 * ln(20+273) + Const) = -0.0107 J/K.

Thus, when mixing 4 kg of water at 80 °C and 6 kg of water at 20 °C, the change in entropy is -0.0107 J/K.

Product description: Digital product "Solving the problem of mixing water of different temperatures"

If you are looking for a high-quality solution to the problem of mixing water of different temperatures, then our digital product is ideal for you! It contains a detailed description of the problem conditions, the formulas and laws used, as well as the derivation of the calculation formula and the answer.

The task is as follows: mix 4 kg of water at 80 °C and 6 kg of water at 20 °C. Our digital product will help you determine the change in entropy during the mixing process.

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This digital product is a detailed solution to the problem of mixing water of different temperatures. The problem combines two portions of water of different temperatures: 4 kg at 80 °C and 6 kg at 20 °C, and it is necessary to determine the change in entropy during the process of mixing them.

To solve the problem, the formula for entropy change is used: ΔS = Skon - Initial, where ΔS is the change in entropy, Skon is the entropy of the final state of the system, Initial is the entropy of the initial state of the system.

For each portion of water, its entropy is found using the formula: S = Cpln(T) + Const, where C is the heat capacity of the substance, T is the temperature in Kelvin, Const is a constant.

When mixing water, the temperature equalizes to an equilibrium state, and the amount of heat transferred from a hotter portion to a colder one can be calculated using the formula: Q = m1 * C1 * (Tcon - Tav), where Q is the amount of heat, m1 is the mass of more hot portion of water, C1 is the heat capacity of water, Tkon is the final temperature of the equilibrium state, Tav is the average temperature of the initial portions of water.

The average temperature of the initial portions of water can be calculated using the formula: Tav = (m1 * T1 + m2 * T2) / (m1 + m2), where m2 is the mass of the colder portion of water, T1 and T2 are the temperatures of the initial portions of water.

The change in entropy can be calculated as the difference between the entropy of the final and initial states: ΔS = Send - Start = (S1 + S2) - Sstart, where S1 and S2 are the entropies of the initial states of two portions of water.

The digital product is presented in a beautiful html format, containing a detailed description of the problem conditions, the formulas and laws used, as well as the output of the calculation formula and the answer. It will help you quickly and accurately solve the problem and get the answer you need.

If you have any questions about the solution, you can ask for help.

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This product is not a physical item, but rather a service in the form of providing a solution to a problem in the field of thermodynamics.

The problem describes the process of mixing two portions of water of different temperatures. To solve the problem, it is necessary to use the laws of thermodynamics, namely the first law of thermodynamics and the law of conservation of energy.

The first step is to determine the change in internal energy of the system, which in this case is a mixture of water at a new temperature. To do this, it is necessary to calculate the amount of heat that transferred from a hot portion of water to a cold portion of water.

The next step is to determine the change in entropy of the system. To do this, it is necessary to use the formula for the change in entropy depending on the change in internal energy and temperature.

After calculating the change in entropy, you can get the answer to the problem. If you have questions about solving a problem, you can seek help from the author of the problem or other specialists in the field of thermodynamics.

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