Question

Определи количество теплоты, необходимое для превращения 4кг льда с температурой 13 градусов в воду при температуре 42 градусов. Температура плавления льда равна (0)градусов, Удельная теплота Плавления льда - 3,4*10 5ДжкгУдельная Теплоемкость льда 2100 Дж/кг*С. Удельная теплоемкость воды - 4200 Дж/кг*С Ответ (округли до десятых) кДж.​

Answer 1

Ответ:

Сначала доводим лёд до температуры плавления 2100×13×4=109200Дж=109.2×10³Дж

Плавим лёд 4×3.4×10⁵=13.6×10⁵Дж

Нагреваем воду 4200×42×4=705.6×10³дж

Складываем 109.2×10³+13.6×10⁵+705.6×10³=828.4×10¹³Дж=828.4×10¹⁰

Объяснение:

Answer 2

Calculation of Heat Required to Convert Ice to Water

To determine the amount of heat required to convert 4 kg of ice at a temperature of 13 degrees Celsius to water at a temperature of 42 degrees Celsius, we need to consider the following factors:

1. Heat required to raise the temperature of ice from -13 degrees Celsius to 0 degrees Celsius. 2. Heat required to melt the ice at 0 degrees Celsius. 3. Heat required to raise the temperature of the resulting water from 0 degrees Celsius to 42 degrees Celsius.

Let's calculate each of these steps separately:

# Heat required to raise the temperature of ice from -13 degrees Celsius to 0 degrees Celsius:

The specific heat capacity of ice is 2100 J/(kg·°C). Therefore, the heat required to raise the temperature of 4 kg of ice by 13 degrees Celsius can be calculated as follows:

Q1 = m × c × ΔT where: - Q1 is the heat required, - m is the mass of the ice (4 kg), - c is the specific heat capacity of ice (2100 J/(kg·°C)), - ΔT is the change in temperature (13 - (-13) = 26 degrees Celsius).

Substituting the values into the equation: Q1 = 4 kg × 2100 J/(kg·°C) × 26 °C = 218,400 J.

# Heat required to melt the ice at 0 degrees Celsius:

The specific latent heat of fusion for ice is 3.4 × 10^5 J/kg. Therefore, the heat required to melt 4 kg of ice can be calculated as follows:

Q2 = m × Lf where: - Q2 is the heat required, - m is the mass of the ice (4 kg), - Lf is the specific latent heat of fusion for ice (3.4 × 10^5 J/kg).

Substituting the values into the equation: Q2 = 4 kg × 3.4 × 10^5 J/kg = 1,360,000 J.

# Heat required to raise the temperature of the resulting water from 0 degrees Celsius to 42 degrees Celsius:

The specific heat capacity of water is 4200 J/(kg·°C). Therefore, the heat required to raise the temperature of 4 kg of water by 42 degrees Celsius can be calculated as follows:

Q3 = m × c × ΔT where: - Q3 is the heat required, - m is the mass of the water (4 kg), - c is the specific heat capacity of water (4200 J/(kg·°C)), - ΔT is the change in temperature (42 - 0 = 42 degrees Celsius).

Substituting the values into the equation: Q3 = 4 kg × 4200 J/(kg·°C) × 42 °C = 705,600 J.

# Total heat required:

To find the total heat required, we sum up the heat required for each step:

Total heat = Q1 + Q2 + Q3 = 218,400 J + 1,360,000 J + 705,600 J = 2,284,000 J

Converting the total heat to kilojoules (kJ): Total heat = 2,284,000 J = 2,284 kJ

Therefore, the amount of heat required to convert 4 kg of ice at a temperature of 13 degrees Celsius to water at a temperature of 42 degrees Celsius is approximately 2,284 kJ (rounded to the nearest tenth).

Please note that the calculations assume no heat loss to the surroundings and that the specific heat capacities and latent heat values provided are accurate for the given substances.

Let me know if there's anything else I can help you with!