Question

Помогите пожалуйста!!! Сколько льда при 0 градусов, может растаять в воде объемом 10 л и температурой 50 градусов???​

Answer 1

С*m*(50-0)=λ*M

4200*10*50=340000*M

M=6,18 кг.

Ответ:6,18 кг.

Answer 2

Calculation of Ice Melting in Water

To calculate the amount of ice that can melt in water, we need to consider the heat transfer between the ice and the water. The heat transfer equation is given by:

Q = m * c * ΔT

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

In this case, we know the volume of water (10 liters) and the initial and final temperatures (0 degrees Celsius and 50 degrees Celsius, respectively). However, we need to convert the volume of water to the mass of ice to use in the equation.

Conversion of Volume to Mass

To convert the volume of water to the mass of ice, we need to know the density of water and the density of ice. The density of water is approximately 1 gram per milliliter, and the density of ice is approximately 0.92 grams per milliliter.

Given that the volume of water is 10 liters, we can calculate the mass of ice using the following equation:

Mass of ice = Volume of water * Density of water - Volume of water * Density of ice

Substituting the values, we get:

Mass of ice = 10,000 grams - 9,200 grams = 800 grams

Now that we have the mass of ice, we can calculate the heat transferred using the heat transfer equation.

Calculation of Heat Transferred

Using the heat transfer equation:

Q = m * c * ΔT

Where: - m = 800 grams (mass of ice) - c = specific heat capacity of ice - ΔT = 50 degrees Celsius - 0 degrees Celsius = 50 degrees Celsius

The specific heat capacity of ice is approximately 2.09 J/g°C Substituting the values, we get:

Q = 800 grams * 2.09 J/g°C * 50 degrees Celsius = 83,600 J

Therefore, the amount of ice that can melt in water with a volume of 10 liters and a temperature of 50 degrees Celsius is approximately 83,600 J.

Please note that this calculation assumes ideal conditions and does not take into account factors such as heat loss to the surroundings or the presence of impurities in the water.