30 g of water at 5 °C was poured into 50 g of water at 90 °C. Definition

30 grams of water at 5 °C were dissolved in 50 grams of water at a temperature of 90 °C. It is necessary to determine the change in the entropy of water.

Solution to problem 20605: Given: m1 = 50 grams (mass of the first portion of water) T1 = 90 °C (temperature of the first portion of water) m2 = 30 grams (mass of the second portion of water) T2 = 5 °C (temperature of the second portion of water)

Find: ΔS (change in entropy of water)

Solution: It is known that the change in entropy of the system is calculated by the formula: ΔS = mc ln (T2/T1) + mc ln (m1 + m2/m1)

where: m - mass of the substance c - specific heat capacity of water (4.18 J/(g °C)) ln - natural logarithm

Substitute the values ​​into the formula: ΔS = 50 g * 4.18 J/(g °C) * ln (278.15 K / 363.15 K) + (50 g + 30 g) * 4.18 J/(g ·°C) * ln (50 g + 30 g / 50 g) ΔS ≈ -0.462 J/K

Answer: ΔS ≈ -0.462 J/K.

From this solution it follows that when two portions of water with different temperatures are mixed, the entropy of the system decreases, which corresponds to an increase in order in the system.

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30 grams of water at 5 °C were dissolved in 50 grams of water at a temperature of 90 °C. It is necessary to determine the change in the entropy of water.

This task can be easily solved using our digital product, which provides calculations using formulas and laws of thermodynamics.

By purchasing our product, you get a unique opportunity to solve problems quickly and efficiently!

The digital product presented in the description can help in solving this problem in the field of thermodynamics.

In the problem, it is known that 30 g of water at 5 °C was dissolved in 50 g of water at a temperature of 90 °C, and it is necessary to determine the change in the entropy of water.

From the formula for calculating the change in entropy of the system, which is given in the solution to problem 20605, it follows that ΔS = mc ln (T2/T1) + mc ln (m1 + m2/m1), where m is the mass of the substance, c is the specific heat capacity of water ( 4.18 J/(g·°C)), ln - natural logarithm, T1 and T2 - temperatures of the first and second portions of water, respectively, m1 and m2 - their masses.

Substituting the known values ​​into the formula, we get: ΔS = 50 g * 4.18 J/(g °C) * ln (278.15 K / 363.15 K) + (50 g + 30 g) * 4.18 J /(g °C) * ln (50 g + 30 g / 50 g) ≈ -0.462 J/K.

Answer: ΔS ≈ -0.462 J/K.

Thus, this digital product can provide calculations using formulas and laws of thermodynamics, which can help solve such problems quickly and efficiently. If you have any questions about solving a problem, do not hesitate to ask for help.

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There is no product description. However, I can help with problem 20605.

The task is to determine the change in the entropy of water when 30 g of water at 5 °C is added to 50 g of water at 90 °C.

To solve the problem it is necessary to use the entropy change formula:

ΔS = Q/T,

where ΔS is the change in entropy, Q is the thermal quantity transferred to the system, T is the temperature of the system in absolute units (Kelvins).

In this problem, the system is a mixture of water at 90 °C and 5 °C after mixing. The thermal amount transferred to the system when water is mixed can be found using the formula:

Q = mcΔT,

where m is the mass of the substance, c is the specific heat capacity of the substance, ΔT is the change in temperature.

The initial entropy of the system is determined using the equation:

S = mCpln(T2/T1),

where S is entropy, m is the mass of the substance, Cp is the specific heat capacity of the substance at constant pressure, T1 is the initial temperature, T2 is the final temperature.

Substituting all known values ​​into the formulas, we get:

Q = (50 g + 30 g) * 4.18 J/(g*°C) * (90 °C - 5 °C) = 11,970 J,

T = (90 °C + 273.15) K = 363.15 K,

ΔS1 = 50 g * 4.18 J/(g*°C) * ln(363.15 K/363.15 K) = 0 J/K,

ΔS2 = 30 g * 4.18 J/(g*°C) * ln(363.15 K/278.15 K) ≈ 24.6 J/K,

ΔS = ΔS1 + ΔS2 = 24.6 J/К.

Answer: The change in entropy of water when mixed under these conditions is approximately 24.6 J/K.

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