Question

4 л воды, взятой при температуре 70 ºС, смешали с водой, температура которой 30 ºС. Определите массу более холодной воды, если известно, что установившаяся в смеси температура равна 40 ºС. Теплообменом с сосудом пренебречь

Answer 1

Cm1∆t1=cm2∆t2
4200Дж/кг×С × 4 кг × 30°С=4200Дж/кг×С × х × 10°С
504000=42000х
х=12,8 кг

Answer 2

Масса холодной воды равна 2 литра при температуре 40

Answer 3

Problem Analysis

We are given 4 liters of water at a temperature of 70ºC and an unknown mass of water at a temperature of 30ºC. We need to determine the mass of the colder water when the final temperature of the mixture is 40ºC. We can assume that there is no heat exchange with the container.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the cold water.

The formula for heat transfer is given by:

Q = mcΔT

Where: - Q is the heat transfer (in joules) - m is the mass of the substance (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)

Since the heat lost by the hot water is equal to the heat gained by the cold water, we can set up the following equation:

m1c1ΔT1 = m2c2ΔT2

Where: - m1 is the mass of the hot water (4 liters) - c1 is the specific heat capacity of water (4.18 J/g°C) - ΔT1 is the change in temperature of the hot water (70°C - 40°C) - m2 is the mass of the cold water (unknown) - c2 is the specific heat capacity of water (4.18 J/g°C) - ΔT2 is the change in temperature of the cold water (40°C - 30°C)

To solve for the mass of the cold water (m2), we can rearrange the equation:

m2 = (m1c1ΔT1) / (c2ΔT2)

Now we can substitute the known values into the equation and calculate the mass of the cold water.

Calculation

Let's calculate the mass of the cold water using the given values:

- m1 = 4 liters = 4000 grams - c1 = 4.18 J/g°C - ΔT1 = 70°C - 40°C = 30°C - c2 = 4.18 J/g°C - ΔT2 = 40°C - 30°C = 10°C

Substituting these values into the equation:

m2 = (4000g * 4.18 J/g°C * 30°C) / (4.18 J/g°C * 10°C)

Simplifying the equation:

m2 = 4000g * 3

Calculating the mass of the cold water:

m2 = 12000g

Therefore, the mass of the colder water is 12000 grams.

Answer

The mass of the colder water, when mixed with 4 liters of water at a temperature of 70ºC, and the resulting temperature is 40ºC, is 12000 grams.