Question

Помогите решить задачи:Сколько энергии необходимо для плавления железного маталолома массой 4 тонны если начальная начальная температура 39 градусов цельсия. В железной коробке массой 300грамм,мальчик расплавил 10 грамм олова .Какое кол-во теплоты пошло на нагревание коробки и плавления олова если начальная температура их была 32 гардуса по цельсию.

Answer 1

Количество теплоты Q= с*m*(t2-t1)
Q железо = 450*0,3*(232-32)=27000 Дж
Q олово = 230*0,1(232-32) = 4600 Дж
Q = 27000+4600=31600 Дж

Answer 2

Calculation of energy required to melt iron scrap:

To calculate the energy required to melt the iron scrap, we need to use the specific heat capacity and the heat of fusion of iron.

The specific heat capacity of iron is approximately 0.45 J/g°C The heat of fusion of iron is approximately 272 J/g.

Given: - Mass of iron scrap = 4 tons = 4,000,000 grams - Initial temperature of iron scrap = 39°C

To calculate the energy required to raise the temperature of the iron scrap from its initial temperature to its melting point, we use the formula:

Q = mcΔT

Where: - Q is the energy required (in joules) - m is the mass of the iron scrap (in grams) - c is the specific heat capacity of iron (in J/g°C) - ΔT is the change in temperature (in °C)

First, let's calculate the energy required to raise the temperature of the iron scrap to its melting point:

Q1 = mcΔT1

Where: - m = 4,000,000 grams - c = 0.45 J/g°C - ΔT1 = (1535°C - 39°C) = 1496°C

Q1 = (4,000,000 g) * (0.45 J/g°C) * (1496°C) = 2,694,000,000 J

Next, let's calculate the energy required to melt the iron scrap:

Q2 = mL

Where: - m = 4,000,000 grams - L = 272 J/g

Q2 = (4,000,000 g) * (272 J/g) = 1,088,000,000 J

Therefore, the total energy required to melt the iron scrap is:

Total energy = Q1 + Q2 = 2,694,000,000 J + 1,088,000,000 J = 3,782,000,000 J

So, the energy required to melt the iron scrap with a mass of 4 tons and an initial temperature of 39°C is approximately 3,782,000,000 joules.

Calculation of heat for heating the iron box and melting the lead:

To calculate the heat required to heat the iron box and melt the lead, we need to use the specific heat capacity and the heat of fusion of iron and lead.

The specific heat capacity of iron is approximately 0.45 J/g°C The specific heat capacity of lead is approximately 0.13 J/g°C The heat of fusion of lead is approximately 24.5 J/g.

Given: - Mass of the iron box = 300 grams - Mass of the lead = 10 grams - Initial temperature of the iron box and lead = 32°C

To calculate the heat required to raise the temperature of the iron box and lead from their initial temperature to their melting point, we use the formula:

Q = mcΔT

Where: - Q is the heat required (in joules) - m is the mass of the substance (in grams) - c is the specific heat capacity of the substance (in J/g°C) - ΔT is the change in temperature (in °C)

First, let's calculate the heat required to raise the temperature of the iron box to its melting point:

Q1 = mcΔT1

Where: - m = 300 grams - c = 0.45 J/g°C - ΔT1 = (1535°C - 32°C) = 1503°C

Q1 = (300 g) * (0.45 J/g°C) * (1503°C) = 203,085 J

Next, let's calculate the heat required to melt the lead:

Q2 = mL

Where: - m = 10 grams - L = 24.5 J/g

Q2 = (10 g) * (24.5 J/g) = 245 J

Therefore, the total heat required to heat the iron box and melt the lead is:

Total heat = Q1 + Q2 = 203,085 J + 245 J = 203,330 J

So, the heat required to heat the iron box and melt the 10 grams of lead with an initial temperature of 32°C is approximately 203,330 joules.