Предмет: Физика, автор: vovanekrasov653

Сколько граммов спирта потребуется, чтобы нагреть до кипения 3 кг воды, взятой при температуре. 20°.

Ответы

Автор ответа: serbakovi458

Ответ:

Q1=cm(t2-t1) Q1=3*4200*(100-20)=1008000 Дж qспирта= 27×106 q*m=Q2 Q1=Q2 m=Q2/q m=1008000/27×106 = 0.0373 кг= 37 г ответ: 37 г.

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