Question

В сосуде 3 литра воды при температуры 20 градусов, опустили кусок железа массой 3 кг нагретый до 540 градусов ,при этом температура воды поднялась до 55 градусов и воды испарилась определите массу воды обратившуюся в пар?

Answer 1

Три умнож  на  два  и получиш  что   испарилось шесть  литров

Answer 2

Problem Analysis

We have a container with 3 liters of water at a temperature of 20 degrees Celsius. A piece of iron weighing 3 kg and heated to 540 degrees Celsius is submerged in the water. As a result, the temperature of the water rises to 55 degrees Celsius, and some of the water evaporates. We need to determine the mass of the water that has turned into steam.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the iron. We can calculate the heat gained by the water using the formula:

Q = m * c * ΔT

Where: - Q is the heat gained by the water - m is the mass of the water - c is the specific heat capacity of water (approximately 4.18 J/g°C) - ΔT is the change in temperature of the water (final temperature - initial temperature)

We can calculate the heat lost by the iron using the formula:

Q = m * c * ΔT

Where: - Q is the heat lost by the iron - m is the mass of the iron - c is the specific heat capacity of iron (approximately 0.45 J/g°C) - ΔT is the change in temperature of the iron (final temperature - initial temperature)

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

m_water * c_water * ΔT_water = m_iron * c_iron * ΔT_iron

We know the initial and final temperatures of the water and the iron, so we can calculate the change in temperature for both substances. We also know the mass of the iron. We can rearrange the equation to solve for the mass of the water:

m_water = (m_iron * c_iron * ΔT_iron) / (c_water * ΔT_water)

Let's calculate the mass of the water that has turned into steam.

Calculation

Given: - Initial temperature of water (T1) = 20 degrees Celsius - Final temperature of water (T2) = 55 degrees Celsius - Initial temperature of iron (T1_iron) = 540 degrees Celsius - Mass of iron (m_iron) = 3 kg

First, let's calculate the change in temperature for both the water and the iron:

ΔT_water = T2 - T1 = 55 - 20 = 35 degrees Celsius ΔT_iron = T1_iron - T1 = 540 - 20 = 520 degrees Celsius

Now, let's calculate the mass of the water:

m_water = (m_iron * c_iron * ΔT_iron) / (c_water * ΔT_water)

Substituting the values:

m_water = (3 kg * 0.45 J/g°C * 520°C) / (4.18 J/g°C * 35°C)

Calculating the mass of the water:

m_water = (3 * 0.45 * 520) / (4.18 * 35) = 0.152 kg

Therefore, the mass of the water that has turned into steam is approximately 0.152 kg.

Answer

The mass of the water that has turned into steam is approximately 0.152 kg.