Предмет: Физика, автор: luerma

Помогите!) Кто что может!

1)Сколько необходимо сжечь спирта, чтобы расплавить лед массой 2 кг, взятый при температуре -5°, а полученную воду нагреть до кипения и 1 кг воды превратить в пар? КПД спиртовки 40%.

2)Сжигая 800 грамм бензина, воду массой 50 кг нагревают от 20 градусов до 100, причём часть воды испаряется. Сколько воды испарилось, если 60% энергии сгорания бензина передано воде?

(Ответы есть, нужны решения)

Ответы

Автор ответа: артём183

теплота сгорания умножить на КПД=теплота нагрево до температуры плаления+теплота плавления

q*m(спирта)*КПД=c*M(чугуна)*(t(плавления)-20)+лямбда*M(чугуна)

m=(c*M*(t(плавления)-20)+лямбда*M)/q*КПД

m=(29,3*(1200-20)*10^12+10^11)/0,4*31·106

m=27882,258*10^5 Дж

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