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A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).

Respuesta :

Answer:

  [tex]c_{e1}[/tex] = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded          Qh = m1 ce1 ([tex]T_{h}[/tex] -[tex]T_{f}[/tex])

Heat absorbed     Qc = m2 ce2 ([tex]T_{f}[/tex] - T₀)

Body 1 is metal and body 2 is water .  Where m are the masses of the two bodies, ce their specific heat and T the temperatures

      Qh = Qc

      m₁ [tex]c_{e1}[/tex] ([tex]T_{h}[/tex]- [tex]T_{f}[/tex]) = m₂  [tex]c_{e2}[/tex] ([tex]T_{f}[/tex] - T₀)

we clear the specific heat of the metal

      [tex]c_{e1}[/tex] = m₂  [tex]c_{e2}[/tex] ([tex]T_{f}[/tex] - T₀) / (m₁ ([tex]T_{h}[/tex]-[tex]T_{f}[/tex]))

     [tex]c_{e1}[/tex]= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

      [tex]c_{e1}[/tex] = 209.2 (9.36) / (75 78.85)

      [tex]c_{e1}[/tex] = 1958.11 / 5913.75

     [tex]c_{e1}[/tex] = 0.331 J / g ° C

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