Question

Определите,какое количество свинца,взятого при 0 градусов Цельсия можно расплавить за счёт теплоты,полученной при сгорании 500 грамм нефти, если КПД нагревателя 80%

Answer 1

T1=0 C
c=0.13 кДЖ/кгград
tпл=327 С
L=24 кДж/кг
m2=0.5 кг
k=44 МДж/кг
n=0.8
m1-?
Q1+Q3=0.8Q2  
Q1=m1c(tпл-t1)  Q3=m1L  
 Q1+Q3=m1(c(tпл-t1)+k)=m1(0.13*327+244000)=m1*66.51 кДж
Q2=km2m1c(tпл-t1)=0.8km2
0.8m2k=m1(c(tпл-t1)+k)
m1=0.8*0.5*44000/66,51=423 кr

Answer 2

Calculation of the Amount of Lead Melted

To determine the amount of lead that can be melted using the heat generated from burning 500 grams of oil at an efficiency of 80%, we need to consider the heat released during the combustion process and the heat required to melt lead.

First, let's calculate the heat released during the combustion of 500 grams of oil. The heat released can be calculated using the formula:

Q = m × c × ΔT

Where: - Q is the heat released - m is the mass of the substance (in this case, oil) - c is the specific heat capacity of the substance - ΔT is the change in temperature

Unfortunately, the specific heat capacity of oil is not provided in the search results. However, we can assume a specific heat capacity of oil to be approximately 2.1 J/g°C based on general values for hydrocarbons.

Using this assumption, we can calculate the heat released during the combustion of 500 grams of oil. Assuming the initial temperature of the oil is 0°C and the final temperature after combustion is 100°C (since the question does not specify the temperature change), we have:

Q = 500 g × 2.1 J/g°C × (100°C - 0°C) = 105,000 J

Now, let's calculate the amount of lead that can be melted using this heat. To do this, we need to know the heat required to melt lead. The heat required to melt a substance can be calculated using the formula:

Q = m × L

Where: - Q is the heat required - m is the mass of the substance (in this case, lead) - L is the latent heat of fusion of the substance

The latent heat of fusion of lead is approximately 24.5 J/g.

Assuming the initial temperature of the lead is 0°C and the final temperature after melting is the melting point of lead, which is 327.5°C, we have:

Q = m × 24.5 J/g × (327.5°C - 0°C) = 7,993.75 J

Now, we can calculate the mass of lead that can be melted using the heat released during the combustion of the oil:

m = Q / L = 7,993.75 J / 24.5 J/g ≈ 326.7 g

Therefore, approximately 326.7 grams of lead can be melted using the heat generated from burning 500 grams of oil at an efficiency of 80%.

Please note that the specific heat capacity of oil and the latent heat of fusion of lead may vary depending on the specific type and purity of the substances used. The values provided here are approximate and should be used as a general guideline.