Question

Сколько литров воды при 83 градусов нужно добавить к 4 л воды при 20 градусов, чтобы получить воду температурой 65 градусов? Теплообменом с окружающей средой пренебречь.

Answer 1

Q1=Q2

 

c*m1*(t3-t2)=c*m2*(t1-t3)

m1*(t3-t2)=m2*(t1-t3)

 

m1=p*V1

 

m2=p*V2

 

p*V1*(t3-t2)=p*V2*(t1-t3)

V1*(t3-t2)=V2*(t1-t3)

 

V2=V1*(t3-t2)/(t1-t3)

 

V2=4* (65-20)/(83-65)

 

V2=4*45/18=10

 

ВСЕ!

 

Answer 2

десять литров ,,,,,,,,,

Answer 3

Calculation of Water Temperature

To calculate the amount of water at 83 degrees Celsius that needs to be added to 4 liters of water at 20 degrees Celsius to obtain water at 65 degrees Celsius, we can use the principle of heat transfer.

The formula for heat transfer is:

Q = mcΔT

Where: - Q is the heat transferred - m is the mass of the substance - c is the specific heat capacity of the substance - ΔT is the change in temperature

In this case, we can assume that the specific heat capacity of water remains constant throughout the temperature range.

Let's break down the calculation step by step:

1. Calculate the heat transferred from the 83-degree water to the final temperature of 65 degrees Celsius. - The mass of the 83-degree water is unknown, so let's assume it is represented by 'm'. - The specific heat capacity of water is approximately 4.18 J/g°C. - The change in temperature is 83 - 65 = 18 degrees Celsius. - Using the formula Q = mcΔT, we can calculate the heat transferred from the 83-degree water to the final temperature: Q1 = m * 4.18 * 18 2. Calculate the heat transferred from the 20-degree water to the final temperature of 65 degrees Celsius. - The mass of the 20-degree water is 4 liters, which is equivalent to 4000 grams. - The specific heat capacity of water is approximately 4.18 J/g°C. - The change in temperature is 65 - 20 = 45 degrees Celsius. - Using the formula Q = mcΔT, we can calculate the heat transferred from the 20-degree water to the final temperature: Q2 = 4000 * 4.18 * 45 3. The total heat transferred is the sum of Q1 and Q2: Q_total = Q1 + Q2

4. Since we are assuming no heat exchange with the surroundings, the total heat transferred should be equal to zero: Q_total = 0

5. Set up the equation and solve for the mass of the 83-degree water (m): Q1 + Q2 = 0 m * 4.18 * 18 + 4000 * 4.18 * 45 = 0

By solving this equation, we can find the mass of the 83-degree water that needs to be added to the 4 liters of water at 20 degrees Celsius to obtain water at 65 degrees Celsius.

Please note that the specific heat capacity of water used in this calculation is an approximation and may vary slightly depending on the temperature range. Additionally, this calculation assumes that the density of water remains constant throughout the temperature range.

Let me know if you need any further assistance!