Respuesta :
Answer:
[tex]c_m =272.365\frac{J}{KgK}[/tex]
Explanation:
1) Notation and data given
[tex]m_c[/tex]= mass of the container = 3.9 kg
[tex]m_w[/tex]= mass of the water inside of the container= 11 kg
[tex]m_m[/tex]=mass of the metal= 2 kg
[tex]T_{im}=189\degree C[/tex] initital temperature of the metal
[tex]T_{iw}=16\degree C[/tex] initital temperature of the water
[tex]T_{ic}=16\degree C[/tex] initital temperature of the container
[tex]T_{f}=18\degree C[/tex] final equilibrium temperature
2) Concepts and formulas
Since the inital temperature for the pice of metal is higher than the temperature for the container and the water inside, the piece of metal will transfer heat, and this transferred heat is the same amount absorbed by the container and the water reaching the equilibrium. So then we have this formula:
[tex]Q_m =Q_w + Q_c[/tex]
We don't have any change of phase so then we just have the presence of sensible heat, and using this we got that:
[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_c \Delta T[/tex]
On this case the container is made by the same metal so then [tex]c_c=c_m[/tex]
[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_m \Delta T[/tex]
And solving for [tex]c_m[/tex] from the last expression we got:
[tex]m_p c_m \Delta T_m -m_c c_m \Delta T_c =m_w c_w \Delta T[/tex]
[tex]c_m [m_p \Delta T_m -m_c \Delta T_c] =m_w c_w \Delta T[/tex]
[tex]c_m=\frac{m_w c_w \Delta T}{m_p \Delta T_m -m_c \Delta T_c}[/tex]
Replacing the values we have:
[tex]c_m=\frac{11kg(4187\frac{J}{KgK}(18\degree C-16\degree C)}{2kg(189\degree C -16 \degree C) -3.9Kg(18\degree C -16\degree C)}= 272.365\frac{J}{Kg K}[/tex]
