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Sagot :
We have in this case a transfer of energy from the metal to the water. If we assume that there are no heat losses to the environment, we can say that the heat released by the metal will be the same heat absorbed by the water, it will have the same magnitude but opposite sign. We can represent this with the following equation.
[tex]Q_m=-Q_w[/tex]The subscript 'm' will be for the metal and its properties and the subscript 'w' for water.
Now, the energy of each object will be:
[tex]m_m.Cp_m.\Delta T_m=-m_w.Cp_w.\Delta T_w[/tex]Where,
m is the mass of the object
Cp is the specific heat of the object
dT is the change of temperature
We clear the mass of the metal
[tex]m_m=\frac{-m_w.Cp_w.\Delta T_w}{Cp_m.\Delta T_m}[/tex]Now, we replace the known data
mw=117g
Cpw=1cal/g°C
dTw=T2-T1=32.3°C-21.3°C=11°C
Cpm=0.104cal/g°C
dTm=T2-T1=32.3°C-95.3°C=-63°C
So, we will have:
[tex]\begin{gathered} m_m=\frac{-117g\times1\frac{cal}{g\degree C}\times11\degree C}{0.104\frac{cal}{g\operatorname{\degree}C}\times(-63\degree C)} \\ m_m=\frac{117\times1\times11}{0.104\times63}g=196g \end{gathered}[/tex]Answer: The mass of the piece of metal will be 196g
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