Предмет: Физика, автор: elizavfedotova

Какое количество теплоты затрачено на расплавление 1 т железа взятого при температуре 39°С? ​


gavrish262190: не то. не пиши это...... пожалуйста
gavrish262190: Дано:                                      Решение: Q=Q1+Q2=cm(t2-t1)+λm
m=1т=1000кг                          Q=460 Дж/кг·°С · 1000кг(1539°С - 39°С) +
t_{1}t1​ =39°C                         +250000 Дж/кг · 1000 кг = 690 кДж + 250 кДж =
t_{2}t2​ =1539°С                     = 940 кДж
λ=2,5·10^5 Дж/кг                     Ответ Q=940 кДж
с=460 Дж/кг·°С
Q-?
gavrish262190: вот правильный

Ответы

Автор ответа: gavrish262190
2

Ответ:

1,095ГДж

Удачи)))))))))))

Приложения:

gavrish262190: Дано:                                      Решение: Q=Q1+Q2=cm(t2-t1)+λm
m=1т=1000кг                          Q=460 Дж/кг·°С · 1000кг(1539°С - 39°С) +
t_{1}t1​ =39°C                         +250000 Дж/кг · 1000 кг = 690 кДж + 250 кДж =
t_{2}t2​ =1539°С                     = 940 кДж
λ=2,5·10^5 Дж/кг                     Ответ Q=940 кДж
с=460 Дж/кг·°С
Q-?
gavrish262190: вот ответ правильный
Автор ответа: dovganryslana
1

Ответ:

m=1т=1000кг - масса железа

t₁=39°C

t₂=1538°С - температура плавления железа

с=450Дж/кг·град - удельная теплоемкость железа

λ=2,7·10⁵ Дж/кг - удельная теплота плавления железа

Решение:

Формула для расчёта кол-ва всей теплоты: Q=Q₁+Q₂

Q₁=cm(t₂-t₁), Q₂=λm,

где Q₁ - энергия, которую нужно затратить, чтобы нагреть железа до температуры плавления, Q₂ - энергия, которую нужно затратить, чтобы железо расплавить.

⇒ E=cm(t₂-t₁)+λm=450*1000*(1538-39)+270000*1000=944550000Дж=944,55МДж

как я понимаю это так)))

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