Tóm tắt :
\(t_1=20^oC\)
\(c=4200J/kg.K\)
\(V_2=20l\rightarrow m_2=D.V=20kg\)
\(t_2=30^oC\)
\(t=95^oC\)
\(m_1=?\)
GIẢI :
Ta có : \(Q_{tỏa}=m_1.c.\left(t-t_1\right)=m_1.4200.\left(95-22\right)\)
\(Q_{thu}=m_2.c.\left(t-t_2\right)=20.4200.\left(95-30\right)=5460000\left(J\right)\)
Ta có : \(Q_{tỏa}=Q_{thu}\)
\(\Rightarrow m_1.c.\left(t-t_1\right)=m_2.c.\left(t-t_2\right)\)
\(\Rightarrow m_1.4200.\left(95-22\right)=20.4200.\left(95-30\right)\)
\(\Rightarrow m_1=\dfrac{5460000}{4200.\left(95-22\right)}\approx17,81kg\)
Vậy lượng nước trong ấm là 17,81kg hay 17,81lít.