Question

Какое количество теплоты нужно предоставить 3,1 кг льда, взятого при температуре −8°С, чтобы его расплавить, а полученную воду нагреть до температуры 100°С кипения и выпарить? Удельная теплоёмкость льда 2,1 кДж/(кг∙К), воды 4,2 кДж/(кг∙К). Удельная теплота плавления льда 332 кДж/кг. Удельная теплота парообразования воды 2,3 МДж/кг. (ответ в МДж значение округлить десятых)

Answer 1

Ответ:

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Answer 2

Calculation of Heat Required to Melt Ice and Heat Water

To calculate the amount of heat required to melt the ice and heat the resulting water to its boiling point, we need to consider the following steps:

1. Heat required to melt the ice: - The specific heat capacity of ice is given as 2.1 kJ/(kg∙K). - The specific latent heat of fusion (melting) of ice is given as 332 kJ/kg. - The mass of the ice is given as 3.1 kg. - The initial temperature of the ice is -8°C.

To calculate the heat required to melt the ice, we can use the formula:

Q1 = m × Lf

Where: - Q1 is the heat required to melt the ice, - m is the mass of the ice, - Lf is the specific latent heat of fusion of ice.

Substituting the given values:

Q1 = 3.1 kg × 332 kJ/kg

Calculating the value of Q1:

Q1 = 1029.2 kJ.

2. Heat required to heat the water to its boiling point: - The specific heat capacity of water is given as 4.2 kJ/(kg∙K). - The mass of the water is the same as the mass of the ice, which is 3.1 kg. - The final temperature of the water is 100°C.

To calculate the heat required to heat the water, we can use the formula:

Q2 = m × Cp × ΔT

Where: - Q2 is the heat required to heat the water, - m is the mass of the water, - Cp is the specific heat capacity of water, - ΔT is the change in temperature.

Substituting the given values:

Q2 = 3.1 kg × 4.2 kJ/(kg∙K) × (100°C - 0°C)

Calculating the value of Q2:

Q2 = 1306.2 kJ.

3. Heat required to vaporize the water: - The specific latent heat of vaporization of water is given as 2.3 MJ/kg.

To calculate the heat required to vaporize the water, we can use the formula:

Q3 = m × Lv

Where: - Q3 is the heat required to vaporize the water, - m is the mass of the water, - Lv is the specific latent heat of vaporization of water.

Substituting the given values:

Q3 = 3.1 kg × 2.3 MJ/kg

Calculating the value of Q3:

Q3 = 7.13 MJ.

Total Heat Required

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

Total Heat = Q1 + Q2 + Q3

Substituting the calculated values:

Total Heat = 1029.2 kJ + 1306.2 kJ + 7.13 MJ

Converting the units to megajoules (MJ):

Total Heat = 1029.2 kJ + 1306.2 kJ + 7.13 MJ = 9.465 MJ

Therefore, the total amount of heat required to melt 3.1 kg of ice taken at -8°C, heat the resulting water to 100°C, and vaporize it is approximately 9.5 MJ (rounded to the nearest tenth)