Once, while I was walking in a park in London, I saw an old strange-looking man. He was sitting on a bench holding a closed book in his hands. I satdown on the bench and looked at the book. I saw that the book was of great interest. It was a very old copy of early Byron's works. I looked at the old man in surprise and undersood that he knew I had sat on the bench because of him and the book he was holding in his hands. I smiled. "It is the last I have," he said and stretched it out to me. I took with the words, "I am a lover of old books."
ответ: 613 680 Дж/кг
Объяснение: сначала запишем
Дано: m= 0.2 то есть переводим 200 грамм в кг 200 делим на 1000
C воды = 4 200 Дж/Кг*° С
tкип. = 100 градусов C
λ лямбда льда = 34*10^4 Дж/кг
L воды = 23*10^5 Дж/кг
(t2-t1)=(100-(-2))
tΔ=100° C
Найти:
Q
сначала нагреем до нуля градусов лёд а потом растопим его
Q1=cm(t2-t1)=4 200 Дж/кг*° C * 0.2кг * 2°= 1 680 Дж/кг
Q2=λm=340 000 Дж/кг *0.2= 68 000 Дж/кг
теперь снова нагреем теперь что бы добраться до 100° и потом испарить
Q3=cm(tΔ)= 4 200 Дж/кг * 0.2 * 100= 84 000 Дж/кг
Q4=Lm= 2 300 000 Дж/кг*0.2= 460 000 Дж/кг
теперь всё сложим и получившиеся значение и будет ответом
Q=Q1+Q2+Q3+Q4= 1 680 Дж/кг+68 000 Дж/кг+84 000 Дж/кг+460 000 Дж/кг = 613 680 Дж/кг