Question

Смесь, сщстоящую из 5 кг льда и 15 кг воды при общей температуре 0С, нужно нагреть до 80С с помощью водяного пара при 100С. Определить необходимое колечество пара. ответ 3.5

Answer 1

Ответ к заданию на рисунке

Answer 2

Problem Analysis

We need to determine the amount of steam required to heat a mixture consisting of 5 kg of ice and 15 kg of water from 0°C to 80°C using steam at 100°C.

Solution

To solve this problem, we need to consider the heat transfer that occurs when the steam condenses and transfers its heat to the mixture of ice and water.

The heat transfer equation can be expressed as follows:

Q = m_water * c_water * ΔT_water + m_ice * c_ice * ΔT_ice + m_steam * h_steam

Where: - Q is the total heat transferred - m_water is the mass of water - c_water is the specific heat capacity of water - ΔT_water is the change in temperature of water - m_ice is the mass of ice - c_ice is the specific heat capacity of ice - ΔT_ice is the change in temperature of ice - m_steam is the mass of steam - h_steam is the enthalpy of steam

We can assume that the specific heat capacity of water is approximately 4.18 kJ/kg°C and the specific heat capacity of ice is approximately 2.09 kJ/kg°C. The enthalpy of steam at 100°C is approximately 2257 kJ/kg.

Calculation

Let's calculate the amount of steam required to heat the mixture.

First, we need to calculate the heat required to heat the water from 0°C to 80°C:

Q_water = m_water * c_water * ΔT_water

Where: - m_water = 15 kg (mass of water) - c_water = 4.18 kJ/kg°C (specific heat capacity of water) - ΔT_water = 80°C - 0°C = 80°C

Q_water = 15 kg * 4.18 kJ/kg°C * 80°C = 5016 kJ

Next, we need to calculate the heat required to melt the ice and then heat it from 0°C to 80°C:

Q_ice = m_ice * c_ice * ΔT_ice

Where: - m_ice = 5 kg (mass of ice) - c_ice = 2.09 kJ/kg°C (specific heat capacity of ice) - ΔT_ice = 0°C - (-80°C) = 80°C

Q_ice = 5 kg * 2.09 kJ/kg°C * 80°C = 836 kJ

Finally, we can calculate the mass of steam required using the equation:

Q = m_steam * h_steam

Where: - Q = Q_water + Q_ice (total heat required) - h_steam = 2257 kJ/kg (enthalpy of steam at 100°C)

m_steam = (Q_water + Q_ice) / h_steam

m_steam = (5016 kJ + 836 kJ) / 2257 kJ/kg = 2.58 kg

Therefore, the required amount of steam is approximately 2.58 kg.

Answer

The required amount of steam to heat the mixture of 5 kg of ice and 15 kg of water from 0°C to 80°C is approximately 2.58 kg.