Cân bằng nhiệt có: \(Q_n=Q_{Cu}\)
\(\Leftrightarrow Q_n=Q_{Cu}=0,5\cdot380\cdot\left(80-20\right)=11400\left(J\right)\)
Ta có: \(Q_n=mc\Delta t\)
\(\Leftrightarrow11400=m\cdot4200\cdot\Delta t\)
\(\Leftrightarrow\Delta t=\dfrac{11400}{4200m}\left(^0C\right)\)
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