Question

Срочно!!! Какое количество теплоты потребуется для нагревания и плавления в железной коробке 100г олова, если их начальная температура была 32 градуса? Масса коробки 30г.

Answer 1

Q1=m1*c1*(t2-t1)+m1*L - нагрев и плавление олова
Q2=m2*c2*(t2-t1) - нагрев коробки
Q=Q1+Q2=m1*c1*(t2-t1)+m1*L+m2*c2*(t2-t1)=
=0,1*218*(232-32)+0,1*60700+0,03*444*(232-32)= 13094 Дж

Answer 2

232 С- температура кипения олова
Q жел=cm(t2-t1)  нагревание железа
Q=460*0.03(232-32)=2 760 Дж  
Q олова=cm(t2-t1)  нагревание олова
Q=250*0.1 (232-32)=5 000 Дж
Q=лm  плавление олова
Q=59 000*0.1=5 900 Дж
Q общ=2 760+5 000+5 900= 13 660 Дж

Answer 3

Calculating Heat Required to Heat and Melt Lead in an Iron Box

To calculate the amount of heat required to heat and melt 100g of lead in an iron box, we can use the formula:

Q = mcΔT + mL

Where: - Q = heat required - m = mass - c = specific heat capacity - ΔT = change in temperature - L = latent heat of fusion

Given: - m_lead = 100g - T_initial = 32°C - m_box = 30g

Let's calculate the heat required step by step.

1. Heat required to raise the temperature of lead from 32°C to its melting point: - The specific heat capacity of lead, c_lead, is 0.128 J/g°C. - The melting point of lead, T_melt, is 327.5°C. - Using the formula Q = mcΔT, where ΔT = T_melt - T_initial: ``` ΔT = 327.5°C - 32°C = 295.5°C Q1 = (100g)(0.128 J/g°C)(295.5°C) = 3849.6 J ```

2. Heat required to melt the lead at its melting point: - The latent heat of fusion of lead, L_lead, is 23.5 J/g. - Using the formula Q = mL: ``` Q2 = (100g)(23.5 J/g) = 2350 J ```

3. Total heat required: ``` Total heat required = Q1 + Q2 = 3849.6 J + 2350 J = 6199.6 J ```

Therefore, the total amount of heat required to heat and melt 100g of lead in an iron box, given that the initial temperature was 32°C, is approximately 6199.6 joules.