[tex]10)Se~da:\\ \\
c_{Al}=919,6\frac {J}{kgK}\\ \\
m_1=500g=0,5kg\\ \\
m_2=3kg\\ \\
c_{apa}=4185\frac{J}{kgK}\\ \\
t_1=20^\circ\\ \\
t_f=100^\circ\\ \\
\eta=40\%=0,4\\ \\
q=4598\times 10^4\frac{J}{kg}\\ \\
m=?g\\ \\ \\
[/tex]
[tex]\eta=\large\frac{Q_u}{Q_t}\\ \\ \\
Q_u=Q_1+Q_2\\ \\
Q_u=c_{Al}\times m_1\times(t_f-t_1)+c_{apa}\times m_2\times(t_f-t_1)\\ \\
Q_u=(t_f-t_1)\times(c_{Al}\times m_1+c_{apa}\times m_2)\\ \\ \\
Q_t=q\times m
[/tex]
[tex]\eta=\frac{(t_f-t_1)\times(c_{Al}\times m_1+c_{apa}\times m_2)}{q\times m}\\ \\
m=\long\frac{(t_f-t_1)\times(c_{Al}\times m_1+c_{apa}\times m_2)}{q\times \eta}\\ \\ \\
Calcule:\\ \\
m=\frac{(100-20)\times(919,6\times 0,5+4185\times 3)}{4598\times 10^4\times 0,4}=0,0566kg=56,6g[/tex]
[tex]11)Se~da:\\ \\
c_{gheta}=2090\frac{J}{kgK}\\ \\
\lambda_{gheata}=33,5\times 10^4\\ \\
m_1=m_2=m=100g=0,1kg\\ \\
t_1=-20^\circ\\ \\
t_2=0^\circ\\ \\
\Delta Q=?\\ \\ \\[/tex]
[tex]Formule:\\ \\
\Delta Q=Q_1-Q_2\\ \\ \\
Q_1=Q_{incalzire}+Q_{topire}\\ \\
Q_1=c_{gheata}\times m\times(t_2-t_1)+m\times \lambda_{gheata}\\ \\ \\
Q_2=Q_{topire}\\ \\
Q_2=m\times \lambda_{gheata}\\ \\ \\[/tex]
[tex]\Delta Q=c_{gheata}\times m\times(t_2-t_1)+m\times \lambda_{gheata}-m\times \lambda_{gheata}\\ \\
\Delta Q=c_{gheata}\times m\times(t_2-t_1)\\ \\ \\
Calcule:\\ \\
\Delta Q=2090\times 0,1\times(0+20)=4180J=4,18kJ
[/tex]