Question

Сколько литров горячей воды при температуре 90°С надо добавить к 10 литрам холодной воды при температуре 10°С, чтобы получить воду,имеющую температуру 40°С? потерями тепла пренебреч.

Answer 1

C×\rho×v1×(t3-t1)=c×\rho×v2×(t3-t2)
10×30=x×50
x=300÷50=6 л

Answer 2

Calculating the Amount of Hot Water Needed

To calculate the amount of hot water needed to be added to 10 liters of cold water at 10°C to achieve a final temperature of 40°C, we can use the principle of mixtures. The formula for this calculation is:

Q = mcΔT

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

Given: - Initial temperature of hot water (T1) = 90°C - Initial temperature of cold water (T2) = 10°C - Final temperature of the mixture (Tf) = 40°C - Mass of cold water (m2) = 10 liters - Specific heat capacity of water (c) = 4.186 J/g°C (or 1 cal/g°C)

We can use the formula to find the mass of hot water needed to achieve the final temperature.

Calculation

Using the formula Q = mcΔT, we can calculate the heat energy required to raise the temperature of the cold water from 10°C to 40°C and the hot water from 90°C to 40°C. Then, we can find the mass of hot water needed.

1. Heat energy to raise the temperature of cold water: - Q2 = m2cΔT2 - Q2 = 10 kg * 4.186 J/g°C * (40°C - 10°C)

2. Heat energy to lower the temperature of hot water: - Q1 = m1cΔT1 - Q1 = m1 * 4.186 J/g°C * (90°C - 40°C)

3. Total heat energy: - Q_total = Q1 + Q2

4. Mass of hot water needed: - m1 = Q_total / (c * (T1 - Tf))

Let's calculate the mass of hot water needed using the given values.

Calculation

1. Heat energy to raise the temperature of cold water: - Q2 = 10 kg * 4.186 J/g°C * (40°C - 10°C) - Q2 = 10 * 4.186 * 30 - Q2 = 1255.8 kJ

2. Heat energy to lower the temperature of hot water: - Q1 = m1 * 4.186 J/g°C * (90°C - 40°C) - Q1 = m1 * 4.186 * 50 - Q1 = 209.3m1 kJ

3. Total heat energy: - Q_total = Q1 + Q2 - Q_total = 1255.8 + 209.3m1

4. Mass of hot water needed: - m1 = Q_total / (c * (T1 - Tf)) - m1 = (1255.8 + 209.3m1) / (4.186 * (90 - 40)) - m1 = (1255.8 + 209.3m1) / 208.6 - 208.6m1 = 1255.8 + 209.3m1 - 208.6m1 - 209.3m1 = 1255.8 - -0.7m1 = 1255.8 - m1 = -1255.8 / -0.7 - m1 ≈ 1794.0 kg

Answer

To achieve a final temperature of 40°C, approximately 1794 liters of hot water at 90°C need to be added to 10 liters of cold water at 10°C, neglecting heat losses.