Question

Сколько необходимо сжечь керосина для превращения 1 кг льда, взятого при температуре -10 °C, а пар при 100 °C? КПД нагревателя 50%?

Answer 1

Оличество тепла от сгорания керосина, которое пойдет на превращение льда = 0,5*46*(10 в 6 степени) *м = 23000000, где м - это масса керосина, которую надо узнать, 46*(10 в 6 степени) - удельная теплота сгорания керосина, 0,5 - потому что 50%.
Лед заберет количество теплоты = 2100*1*10 + 340000*1 + 4200*1*100 + 2256000*1 = 3037000 Дж
Тут: 2100 - удельная теплоемкость льда, 340000 - удельная теплота плавления льда, 4200 - удельная теплоемкость воды, 2256000 - удельная теплота парообразования воды.
Теперь приравняем первое и второе и выразим м (лучше сначала записать это в формулах, а потом только подставлять) . м = 3037000/23000000 = 0,13 кг керосина надо сжечь.

Answer 2

Calculating the Amount of Kerosene Needed to Convert Ice to Steam

To calculate the amount of kerosene needed to convert 1 kg of ice at -10°C to steam at 100°C with an efficiency of 50%, we can use the specific heat capacities of ice, water, and steam, along with the latent heat of fusion and vaporization.

The specific heat capacity of ice (c_ice) is 2.09 J/g°C, the specific heat capacity of water (c_water) is 4.18 J/g°C, and the specific heat capacity of steam (c_steam) is 2.03 J/g°C. The latent heat of fusion of ice (L_fusion) is 334 J/g, and the latent heat of vaporization of water (L_vaporization) is 2260 J/g.

Using these values, we can calculate the amount of kerosene needed.

1. Heating the Ice from -10°C to 0°C: - The energy required to heat the ice from -10°C to 0°C can be calculated using the formula: Q = m * c_ice * ΔT, where m is the mass of the ice and ΔT is the change in temperature. - Substituting the values, we get: Q_ice = 1 kg * 2.09 J/g°C * 10°C = 20.9 kJ.

2. Melting the Ice at 0°C: - The energy required to melt the ice at 0°C can be calculated using the formula: Q = m * L_fusion, where L_fusion is the latent heat of fusion. - Substituting the values, we get: Q_fusion = 1 kg * 334 J/g = 334 kJ.

3. Heating the Water from 0°C to 100°C: - The energy required to heat the water from 0°C to 100°C can be calculated using the formula: Q = m * c_water * ΔT. - Substituting the values, we get: Q_water = 1 kg * 4.18 J/g°C * 100°C = 418 kJ.

4. Vaporizing the Water at 100°C: - The energy required to vaporize the water at 100°C can be calculated using the formula: Q = m * L_vaporization. - Substituting the values, we get: Q_vaporization = 1 kg * 2260 J/g = 2260 kJ.

5. Total Energy Required: - Adding up the energy for each step, we get: Total energy = Q_ice + Q_fusion + Q_water + Q_vaporization = 20.9 kJ + 334 kJ + 418 kJ + 2260 kJ = 3032.9 kJ.

6. Accounting for Efficiency: - Since the efficiency of the heater is 50%, we need to account for this in our calculation. - The actual energy required = Total energy / Efficiency = 3032.9 kJ / 0.50 = 6065.8 kJ.

Therefore, approximately 6065.8 kJ of energy would be required to convert 1 kg of ice at -10°C to steam at 100°C with a heater efficiency of 50%.