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A 1.00 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at 18.0°C. After gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of 22.0°C . The specific heat capacity of water is cw = 4190 J/(kg * K) .
1. Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat capacity of the metal cm?
2. Find the temperature change of the water, ΔTw, and that of the metal, ΔTm.Express your answers in kelvins separated by a comma.

Respuesta :

Answer:

1)  ce₁ = 214.87 J / kg K

,  2)  ΔTw =  4K

,   ΔTm =  78K

Explanation:

1) This is a calorimetry problem, as the anime vessel does not absorb energy, all the energy transferred by the metal and equal to the energy absorbed by the water  

      Qc = - Qa  

      m₁ ce₁ (T₁ - [tex]T_{f}[/tex] ) = m₂ ce₂ ([tex]T_{f}[/tex]-T₀)  

Where index 1 and 2 represents metal and water,

      ce₁ = m₂ / m₁ ce₂ ([tex]T_{f}[/tex]-T₀) / (T₁ - [tex]T_{f}[/tex])  

Let's reduce the temperature to the SI system  

      T₁ = 100 + 273 = 373K  

      T₀ = 18 + 273 = 291K  

      [tex]T_{f}[/tex] = 22 + 273 = 295K  

Let's calculate  

      ce₁ = 1/1 4190 (295 - 291) / (373 - 295)  

      ce₁ = 4190 4/78  

      ce₁ = 214.87 J / kg K  

2)

        ΔTw = 295-291 = 4K

       Δ Tm = 373-295 = 78K

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