Dạng bài như này em thuộc công thức là làm được hén

Ta có: \(\left\{{}\begin{matrix}Q_1=0,3\cdot380\cdot\left(100-15\right)=22440\left(J\right)\\Q_2=1\cdot4200\cdot\left(100-15\right)=357000\left(J\right)\end{matrix}\right.\)

\(\Rightarrow Q=Q_1+Q_2=22440+357000=379440\left(J\right)\)

a.

Cân bằng nhiệt:

 \(Q_{thu}=Q_{toa}=mc\left(t_1-t\right)=0,15\cdot880\cdot\left(100-25\right)=9900\left(J\right)\)

b. 

Ta có: \(Q_{thu}=mc\left(t-t_1\right)\)

\(\Leftrightarrow9900=m\cdot4200\cdot\left(25-20\right)\)

\(\Leftrightarrow m\approx0,5\left(kg\right)\)

Ta có: \(m=DV\Rightarrow V=\dfrac{m}{D}=\dfrac{0,5}{1000}=5\cdot10^{-4}\left(m^3\right)\)

Nhiệt lượng toả ra:

\(Q_{toa}=mc\left(t_1-t_2\right)=2,5\cdot460\cdot\left(150-50\right)=115000\left(J\right)\)

a.

\(Q_{toa}=mc\left(t_1-t\right)=0,2\cdot880\cdot\left(100-27\right)=12848\left(J\right)\)

b.

Cân bằng nhiệt: \(Q_{thu}=Q_{toa}\)

\(\Leftrightarrow m\cdot4200\cdot\left(27-20\right)=12848\)

\(\Leftrightarrow m\approx0,44\left(kg\right)\)

a.

\(Q=mc\left(t_2-t_1\right)=0,6\cdot460\cdot\left(100-20\right)=22080\left(J\right)\)

b.

Cân bằng nhiệt: \(Q_{thu}=Q_{toa}\)

\(\Leftrightarrow m\cdot4200\cdot\left(40-30\right)=0,6\cdot460\cdot\left(100-40\right)\)

\(\Leftrightarrow42000m=16560\)

\(\Leftrightarrow m\approx0,4\left(kg\right)\)

a.

Cân bằng nhiệt:

 \(Q_{thu}=Q_{toa}=0,2\cdot380\cdot\left(100-30\right)=5320\left(J\right)\)

b.

Ta có: \(Q_{thu}=mc\Delta t\)

\(\Leftrightarrow5320=0,2\cdot4200\cdot\Delta t\)

\(\Leftrightarrow\Delta t\approx6,3^0C\)

Ta có: \(\left\{{}\begin{matrix}Q_1=0,4\cdot880\cdot\left(100-24\right)=26752\left(J\right)\\Q_2=1\cdot4200\cdot\left(100-24\right)=319200\left(J\right)\end{matrix}\right.\)

\(\Rightarrow Q=Q_1+Q_2=26752+319200=345952\left(J\right)\)

a.

\(A=Fs=30\cdot50=1500\left(J\right)\)

b. 

\(H=\dfrac{A'}{A''}100\%=\dfrac{55\cdot10\cdot5}{1500+\left(55\cdot10\cdot5\right)}100\%=64,7\%\)

a.

\(P=\dfrac{A}{t}\Rightarrow A=Pt=1800\cdot20=36000\left(J\right)\)

b. 

\(H=\dfrac{A_1}{A_2}100\%=\dfrac{250\cdot10\cdot12}{36000}100\%=\dfrac{30000}{36000}100\%\approx83,3\%\)