Перед сжатием давления газа в цилиндре двигателя внутреннего сгорания равно 8*10^4 па, а температура 50 градусов с.определите температуру газа в конце такта сжатия, если при этом объем его уменьшился в 5 раз, а давление увеличилось до 8*10^5 па.

Лови ,...............

Abdilda Tazhibayev was born in 1909 in Akmeshit (now Kyzylorda). Medium

one hundred

The Kipchak tribe Tory clan is descended from Bessary's grandfather. [1] [2] His father, Tazhibai, died in 1915 and was left in the care of his mother, Aimankul. Due to his condition, six-year-old Abdilda was taken to a suburb of Karaozek. Then the future poet lived until the age of thirteen with his mother Aimankul and uncle Daldabay

grows in upbringing. Young Abdilda's poems and literature

His mother and aunt also aroused his curiosity

will be the grandfather. Aimankul is fluent in Arabic and Persian

He was literate, open-minded, and cheerful.

He writes letters to Abdilda and teaches her literature. Abdilda's mother Aimankul, who later became a well-known folk poet and published a collection of poems, at the request of his father Daldabay, recited to the villagers the stories of the East and various Kazakh poems. Passionate, silent, silent

One of his listeners was Abdil. Later, in a small group, Abdilda would recite the stories and epics to his guests and villagers.

ответ:нагревание льда 2 кг

Q1=c1m1*dT=2100*2*(0-(-5))=21 кДж

плавление льда 2 кг

Q2=λm1 =330 кДж/кг * 2 кг = 660 кДж

нагревание воды 1 кг

Q3 = c2m2*dT =4200*1*(100-0)=420 кДж

испарение воды 1 кг

Q4= Lm2 =2260 кДж\кг * 1 кг = 2260 кДж

полезная энергия Q=Q1+Q2+Q3+Q4=21+660+420+2260 =3361 кДж

КПД спиртовки 40% (0.4)

полная энергия при сгорании W=Q/КПД =3361/0.4 =8402,5 кДж=8402,5 *10^3 Дж

масса спирта m=W/q =8402,5 *10^3 / (2.7*10^7)=0.3112 кг=311.2 г

ОТВЕТ 0.3112 кг=311.2 г

Q=c льда mt+Lm+с воды mt Q= 2100*0.1*5 + 3,3*10^5 *0.1 + 4200*0.1*20 = 1050 +33000+8400=42 450 Дж

Q1=c1*m1*Δt1=460*2*(500-24)=437 920 Дж

Теперь посмотрим какое количество теплоты забрала вся вода для нагревания до 24 градусов:

Q2=c2*m2*Δt2=4200*10*(24-20)=168 000Дж

ΔQ=437 920-168 000=269 920 Дж - теплота, ушедшая на нагревание некоторой воды до 100 градусов и испарения:

Q3=c2*m3*Δt3+L2*m3

m3=Q3/(c2*Δt3+L2)=269 920/(4200*(100-24)+2 300 000)=0.103 кг или 103 г

ответ: испарилось 103 г воды