Question

В холодную воду, массой 0,8 кг долили кипяток массой 0,2 кг. Температура данной смеси =40°С. Определите температуру холодной воды до её нагревания тепловыми потерями пренебречь.

Answer 1

В холодную воду, массой m1=0,8 кг долили кипяток е2=100 С массой m2= 0,2 кг. Температура данной смеси t =40°С. Определите температуру холодной воды t1 -? до её нагревания тепловыми потерями пренебречь.
Q1=Q2
c*m1*(t-t1) = c*m2(t2-t)
0,8*(40-t1) =0,2*60

40  - t1 = 15
t1=40-15=25 C
Ответ t1=25 C

Answer 2

0,2×4200×60=0,8×4200×x x=15 x= Tконечное - Tначальное Тначальное = 40-15=25 град. Цельсия

Answer 3

Problem Analysis

We are given a mixture of cold water and hot water. The mass of the cold water is 0.8 kg, the mass of the hot water (boiling water) is 0.2 kg, and the temperature of the mixture is 40°C. We need to determine the temperature of the cold water before it was heated, assuming no heat loss.

Solution

To solve this problem, we can use the principle of conservation of energy. The total energy of the system before and after heating should be the same.

The energy of an object can be calculated using the formula:

Q = m * c * ΔT

Where: - Q is the energy (in joules) - m is the mass of the object (in kilograms) - c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

In this case, we can assume that the specific heat capacity of water is constant and equal to 4.18 J/g°C.

Let's calculate the energy of the cold water before heating and the energy of the hot water before mixing.

The energy of the cold water before heating can be calculated as:

Q1 = m1 * c * ΔT1

Where: - m1 is the mass of the cold water (0.8 kg) - ΔT1 is the change in temperature of the cold water (unknown)

The energy of the hot water before mixing can be calculated as:

Q2 = m2 * c * ΔT2

Where: - m2 is the mass of the hot water (0.2 kg) - ΔT2 is the change in temperature of the hot water (100°C - unknown)

Since the total energy before and after mixing should be the same, we can equate the two equations:

Q1 + Q2 = 0

Substituting the values, we get:

(m1 * c * ΔT1) + (m2 * c * ΔT2) = 0

Simplifying the equation, we have:

m1 * ΔT1 + m2 * ΔT2 = 0

Now we can solve for ΔT1, which represents the change in temperature of the cold water:

ΔT1 = (-m2 * ΔT2) / m1

Substituting the given values, we get:

ΔT1 = (-0.2 kg * 100°C) / 0.8 kg

Simplifying the equation, we have:

ΔT1 = -25°C

Since we are looking for the temperature of the cold water before heating, we can calculate it by subtracting the change in temperature from the final temperature of the mixture:

T1 = T2 - ΔT1

Substituting the given values, we get:

T1 = 40°C - (-25°C)

Simplifying the equation, we have:

T1 = 65°C

Therefore, the temperature of the cold water before heating was 65°C.

Answer

The temperature of the cold water before it was heated, neglecting heat loss, is 65°C.