Question

Answer 1

Bài 2:

\(Q_{toa}=A=I^2Rt=2^2\cdot220\cdot5\cdot60=264000J\)

\(H=\dfrac{Q_{thu}}{Q_{toa}}1005=>Q_{thu}=Q_{toa}\cdot H\cdot100\%=264000\cdot90\cdot100=2376000000J\)

\(Q_{thu}=mc\Delta t=>m=\dfrac{Q_{thu}}{c\Delta t}=\dfrac{2376000000}{4200\cdot75}\approx7542,8\left(kg\right)=7542,8\left(l\right)\)

\(A=I^2Rt=2^2\cdot220\cdot30=26400\)Wh = 26,4kWh

\(=>T=A\cdot1500=26,4\cdot1500=39600\left(dong\right)\)

Answer 2

Bài 1.

Để hđ bình thường: \(U_m=U_{đm}=220V\)

\(I=\dfrac{P}{U}=\dfrac{100}{220}=\dfrac{5}{11}A\)

\(A=UIt=220\cdot\dfrac{5}{11}\cdot2\cdot3600=720000J=0,2kWh\)

Answer 3

Bài 1:

a. Mắc nối tiếp và HĐT = 220V. \(I=\dfrac{P}{U}=\dfrac{100}{220}=\dfrac{5}{11}A\)

b. \(\left\{{}\begin{matrix}A=Pt=100\cdot2\cdot3600=720000J\\A=Pt=100\cdot2=200Wh=0,2kWh\end{matrix}\right.\)

Answer 4

Bài 2.

\(Q_{tỏa}=R\cdot I^2t=220\cdot2^2\cdot5\cdot60=264000J\)

\(A=UIt=RI^2t=220\cdot2^2\cdot20\cdot60=1056000J\)

\(H=90\%\Rightarrow Q_i=1056000\cdot90\%=950400J\)

Mà \(Q_i=mc\Delta t=m\cdot4200\cdot\left(100-25\right)=950400\)

\(\Rightarrow m=3,02kg\Rightarrow V=3,02l\)

\(A=UIt'=RI^2t'=220\cdot2^2\cdot30\cdot3600=95040000J=26,4kWh\)

\(T=26,4\cdot1500=39600\left(đồng\right)\)