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Brass is an alloy made from copper and zinc. A 0.59 kg brass sample at 90.0 C is dropped into 2.80 kg of water at 5.0 C. If the equilibrium temperature is 6.8C, what is the specific heat capacity of brass?

Respuesta :

 Since no heat is lost to the surroundings, the change in thermal energy of the brass piece and the water sample have have the same magnitude: 

m_brass ∙ c_brass ∙ ∆T_brass = m_water ∙ c_water ∙ ∆T_water 

=> 

c_brass = c_water ∙ (m_water/m_brass) ∙ (∆T_water/∆T_brass) 
= 4186J/kg°C ∙ (2.8kg/0.59kg) ∙ ((6.8°C - 5°C) / (98°C - 6.8°C)) 
= 392J/kg 

The specific heat capacity of brass will be "392 J/Kg".

According to the question,

  • Mass of brass, [tex]m_{brass} = 0.59 \ kg[/tex]
  • Mass of water, [tex]m_{water} = 2.80 \ kg[/tex]

As we know,

→ [tex]m_{brass}\times c_{brass}\times \Delta T_{brass} = m_{water}\times c_{water}\times \Delta T_{water}[/tex]

or,

→ [tex]c_{brass} = c_{water}\times (\frac{m_{water}}{m_{brass}} )\times (\frac{\Delta T_{water}}{\Delta T_{brass}} )[/tex]

By substituting the above values, we get

            [tex]= 4186\times \frac{2.8}{0.59}\times (\frac{6.8-5}{98^{\circ} C-6.8^{\circ} C} )[/tex]

            [tex]= 392 \ J/Kg[/tex]

Thus the above answer is right.

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