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Консультация № 196645
11.10.2019, 03:40
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Здравствуйте! У меня возникли сложности с таким вопросом:
376.Чайник имеет две секции нагревательной проволоки. Время доведения до кипения определенного количества воды после включения первой или второй секции равно T1 =10 мин, T2 =20 мин. Найти время закипания того же количества воды, имеющей ту же начальную температуру после одновременного включения обеих секций, если они соединены: а) параллельно, б) последовательно. К.П.Д. чайника во всех случаях одинаков.
376.Чайник имеет две секции нагревательной проволоки. Время доведения до кипения определенного количества воды после включения первой или второй секции равно T1 =10 мин, T2 =20 мин. Найти время закипания того же количества воды, имеющей ту же начальную температуру после одновременного включения обеих секций, если они соединены: а) параллельно, б) последовательно. К.П.Д. чайника во всех случаях одинаков.
Обсуждение
13.10.2019, 19:22
общий
это ответ
Здравствуйте, dar777!
Дано: T1=10 мин -- время доведения до кипения определённого количества воды после включения первой секции нагревательной проволоки; T2=20 мин -- время доведения до кипения того же количества воды после включения второй секции.
Определить: а) T' -- время доведения до кипения того же количества воды после включения обеих секций, если они соединены параллельно; б) T'' -- время доведения до кипения того же количества воды, если секции соединены последовательно.
Решение
Обозначим R1, R2 -- соответственно сопротивления первой и второй секций чайника; U -- напряжение в сети; Q=U2T/R -- количество теплоты, которое требуется, чтобы довести рассматриваемое количество воды до кипения; T -- время доведения до кипения; R -- сопротивление подключенных к источнику напряжения секций. Согласно условию задачи Q=Q1=U2T1/R1, Q=Q2=U2T2/R2, T1/R1=T2/R2, R2=T2R1/T1.
1. Пусть секции соединены параллельно. Тогда для общего сопротивления R' секций имеем 1/R=1/R1+1/R2=(R1+R2)/(R1R2), R'=R1R2/(R1+R2). Количество теплоты Q=Q'=U2T'/R', откуда T'/R'=T1/R1, T'=T1R'/R1=(T1/R1)(R1R2/(R1+R2))=T1R2/(R1+R2)=(T1(T2R1/T1))/(R1+T2R1/T1)=(T2R1)/(R1(T1+T2)/T1)=T1T2/(T1+T2), то есть T'=10*20/(10+20)=200/30=20/3 (мин).
2. Пусть секции соединены последовательно. Тогда общее сопротивление секций R''=R1+R2. Количество теплоты Q''=U2T''/R'', откуда T''/R''=T1/R1, T''=T1R''/R1=T1(R1+R2)/R1=T1(1+R2/R1)=T1(1+(T2R1/T1)/R1)=T1(1+T2/T1)=T1+T2, то есть T''=10+20=30 (мин).
Ответ: а) 20/3=6,(6)[$8776$]6,67 мин; б) 30 мин.
Дано: T1=10 мин -- время доведения до кипения определённого количества воды после включения первой секции нагревательной проволоки; T2=20 мин -- время доведения до кипения того же количества воды после включения второй секции.
Определить: а) T' -- время доведения до кипения того же количества воды после включения обеих секций, если они соединены параллельно; б) T'' -- время доведения до кипения того же количества воды, если секции соединены последовательно.
Решение
Обозначим R1, R2 -- соответственно сопротивления первой и второй секций чайника; U -- напряжение в сети; Q=U2T/R -- количество теплоты, которое требуется, чтобы довести рассматриваемое количество воды до кипения; T -- время доведения до кипения; R -- сопротивление подключенных к источнику напряжения секций. Согласно условию задачи Q=Q1=U2T1/R1, Q=Q2=U2T2/R2, T1/R1=T2/R2, R2=T2R1/T1.
1. Пусть секции соединены параллельно. Тогда для общего сопротивления R' секций имеем 1/R=1/R1+1/R2=(R1+R2)/(R1R2), R'=R1R2/(R1+R2). Количество теплоты Q=Q'=U2T'/R', откуда T'/R'=T1/R1, T'=T1R'/R1=(T1/R1)(R1R2/(R1+R2))=T1R2/(R1+R2)=(T1(T2R1/T1))/(R1+T2R1/T1)=(T2R1)/(R1(T1+T2)/T1)=T1T2/(T1+T2), то есть T'=10*20/(10+20)=200/30=20/3 (мин).
2. Пусть секции соединены последовательно. Тогда общее сопротивление секций R''=R1+R2. Количество теплоты Q''=U2T''/R'', откуда T''/R''=T1/R1, T''=T1R''/R1=T1(R1+R2)/R1=T1(1+R2/R1)=T1(1+(T2R1/T1)/R1)=T1(1+T2/T1)=T1+T2, то есть T''=10+20=30 (мин).
Ответ: а) 20/3=6,(6)[$8776$]6,67 мин; б) 30 мин.
5
Это самое лучшее решение!
Об авторе:
Facta loquuntur.
Facta loquuntur.
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