Question

С 1. Какое количество горячей воды с температурой 800С нужно налить в холодную воду массой 20 кг и температурой 100С, чтобы установилась температура смеси 300С. Потерями энергии пренебречь

Answer 1

T1=80 C
m2=20 кг
t2=10 C
T=30 C
m1-?
Q1=Q2
m1c(t1-T)=m2c(T-t2)   m1=m2(T-t2)/(t1-T)
m1=20*20/50=8кг

Answer 2

Calculation of the Amount of Hot Water Needed

To determine the amount of hot water with a temperature of 80°C needed to mix with cold water weighing 20 kg and having a temperature of 10°C to achieve a final temperature of 30°C, we can use the principle of conservation of energy.

The equation for the conservation of energy is:

m1 * c1 * (Tf - T1) = m2 * c2 * (T2 - Tf)

Where: - m1 is the mass of the hot water - c1 is the specific heat capacity of water - Tf is the final temperature of the mixture (30°C) - T1 is the initial temperature of the hot water (80°C) - m2 is the mass of the cold water (20 kg) - c2 is the specific heat capacity of water - T2 is the initial temperature of the cold water (10°C)

Since the specific heat capacity of water is approximately 4.2 J/g°C, we can substitute the values into the equation and solve for m1.

m1 * 4.2 * (30 - 80) = 20 * 4.2 * (10 - 30)

Simplifying the equation:

m1 * (-210) = 20 * (-420)

Dividing both sides by -210:

m1 = 20 * (-420) / (-210)

Calculating the value:

m1 = 40 kg

Therefore, you would need 40 kg of hot water with a temperature of 80°C to mix with the cold water to achieve a final temperature of 30°C.