She became the first woman to be awarded this title in the great patriotic war.about the feat zoe kosmodemyanskaya already in 1944, he made the film, the monuments, written poems and named in her honor hundreds of streets.the victory in the great patriotic war - the feat and the glory of our nation. zoya a. kosmodemyanskaya ("tanya") - hero of the soviet union. zoya kosmodemyanskaya was born september 8, 1923 in the village in tambov region. in school she was very ill , but was a good student and was very fair . 31 october 1941 zoe voluntarily became a fighter. the training was very short - 4 november zoe threw in city volokolamsk.she successfully overcame all the difficulties. november 18, 1941 part of zoe received a combat mission - destroy 10 settlements.in part one of the groups went on the job and zoya.at the village golovko group zoe came under fire and collapsed.she alone burned homes.zoe was taken prisoner , over her terribly tortured by the germans, but she was brave.29 november 1941, zoya kosmodemyanskaya was hanged . 27 january 1942 about feat zoe kosmodemyanskaya was the first article in the press.
Объяснение:
Дано:
Q1=80МДж
m2=20кг
q2=27МДж
η-?
η=Q1/Q2=Q1/(m2*q2)=80/(20*27)=0,15
ответ: η=0,15
Дано:
m1=0,5кг
m2=20г=0,02кг
t2=50°C
L=2300000Дж/кг
с=4200Дж/кг°С
t1-?
Теплота от пара:
Q=L*m=2300000*0,02=46000Дж
Теплота для нагрева воды находим по формуле:
Q=c*m*Δt
Отсюда Δt=Q/(m*c)=46000/(4200*0,5)=21,9°С
t1=t2-Δt=50-21,9=28,1°С
ответ: начальная температура воды 28,1°С
Формула для нахождения количества теплоты для превращения жидкости в пар имеет следующий вид:
Q=L*m
Дано:
m=2кг
λ=330кДж/кг
Q-?
Q=λm=330*2=660кДж
ответ: для плавления 2 кг льда нужно 660кДж теплоты.