Предмет: Физика, автор: fedorovakaterin

какое количество теплоты потребует для нагревания 1,2 т. льда от -20 С до температуры плавления , если удельная теплоёмкость льда 2100 Дж:кг С

Ответы

Автор ответа: RuMiro

Q=(лямбда)*m
m=1.2т=1200кг
Q=2100дж/кг*С*1200кг=2520000Дж

InvisorTech: Первая формула - неверная. Впрочем как и сам ответ.

RuMiro: Да ,я извиняюсь.

RuMiro: Отметь мой ответ как "Нарушение " Чтобы его аннулировали .

InvisorTech: Я не отмечаю нарушения. Если модератор придет, отправит на исправление.

RuMiro: Спасибо. Удачного времяпровождения

Автор ответа: InvisorTech

$\boxed{Q = cm\Delta t }\\ \\ Q = 2100 * 1200 * 20 = 50400000 \ [Ds] = 5.04 * 10^{7} \[Ds]$
Ответ: Q = 5.04 × 10⁷ Дж

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