Question

Смесь, состоящую из некоторого количества льда и 2 кг воды при температуре 0°С, нужно нагреть до температуры 80°С с помощью водяного пара массой 3,5 кг при 100°C, 25% которого уходит в окружающую среду. Найти массу льда в первоначальной смеси. Какова будет масса воды после установления теплового равновесия?

Answer 1

Ответ:

1084.18 02х20209 8.85 и 8

Answer 2

Problem Analysis

We are given a mixture consisting of some amount of ice and 2 kg of water at a temperature of 0°C. We need to heat this mixture to a temperature of 80°C using steam with a mass of 3.5 kg at 100°C, 25% of which escapes into the surrounding environment. We need to find the mass of ice in the initial mixture and the mass of water after thermal equilibrium is reached.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat gained by the ice and water mixture should be equal to the heat lost by the steam.

Let's denote the mass of ice in the initial mixture as m_ice and the mass of water as m_water.

The heat gained by the ice and water mixture can be calculated using the formula:

Q_gain = (m_ice * L_f) + (m_water * c * ΔT)

Where: - L_f is the latent heat of fusion of ice (334 kJ/kg) - c is the specific heat capacity of water (4.18 kJ/kg°C) - ΔT is the change in temperature (80°C - 0°C = 80°C)

The heat lost by the steam can be calculated using the formula:

Q_loss = (0.75 * m_steam * c_steam * ΔT_steam) + (0.25 * m_steam * L_v)

Where: - m_steam is the mass of steam (3.5 kg) - c_steam is the specific heat capacity of steam (2.03 kJ/kg°C) - ΔT_steam is the change in temperature of the steam (100°C - 80°C = 20°C) - L_v is the latent heat of vaporization of steam (2260 kJ/kg)

Since the heat gained by the ice and water mixture is equal to the heat lost by the steam, we can set up the equation:

(m_ice * L_f) + (m_water * c * ΔT) = (0.75 * m_steam * c_steam * ΔT_steam) + (0.25 * m_steam * L_v)

We can rearrange this equation to solve for m_ice:

m_ice = ((0.75 * m_steam * c_steam * ΔT_steam) + (0.25 * m_steam * L_v) - (m_water * c * ΔT)) / L_f

To find the mass of water after thermal equilibrium is reached, we can subtract the mass of ice from the initial mass of water:

m_water_final = m_water - m_ice

Let's calculate the values using the given data:

m_steam = 3.5 kg c_steam = 2.03 kJ/kg°C ΔT_steam = 20°C L_v = 2260 kJ/kg m_water = 2 kg c = 4.18 kJ/kg°C ΔT = 80°C L_f = 334 kJ/kg

Substituting these values into the equations, we can calculate the mass of ice and the mass of water after thermal equilibrium is reached.

Calculation

m_ice = ((0.75 * 3.5 * 2.03 * 20) + (0.25 * 3.5 * 2260) - (2 * 4.18 * 80)) / 334

m_water_final = 2 - m_ice

Let's calculate the values:

m_ice = ((0.75 * 3.5 * 2.03 * 20) + (0.25 * 3.5 * 2260) - (2 * 4.18 * 80)) / 334 = 0.147 kg

m_water_final = 2 - 0.147 = 1.853 kg

Answer

The mass of ice in the initial mixture is approximately 0.147 kg. The mass of water after thermal equilibrium is reached is approximately 1.853 kg.

Please note that the values calculated are approximate due to rounding off during the calculations.