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A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of water at a temperature of 23.0°C. After stirring, the final temperature of both copper and water is 26.0°C. Assuming no heat losses, and that the specific heat capacity of water is 4.184 J/(g·°C), what is the molar heat capacity of aluminum, Cm(Al)?

Respuesta :

Answer: The molar heat capacity of aluminum is [tex]25.3J/mol^0C[/tex]

Explanation:

[tex]heat_{absorbed}=heat_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]         .................(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of water = 130.0 g

[tex]m_2[/tex] = mass of aluminiunm = 23.5 g

[tex]T_{final}[/tex] = final temperature = [tex]26.0^oC=(273+26)K=299K[/tex]

[tex]T_1[/tex] = temperature of water = [tex]23^oC=(273+23)K=296K[/tex]

[tex]T_2[/tex] = temperature of aluminium = [tex]100^oC=273+100=373K[/tex]

[tex]c_1[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

[tex]130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)][/tex]

[tex]c_2=0.938J/g^0C[/tex]

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =[tex]0.938J/g^0C\times 27g/mol=25.3J/mol^0C[/tex]

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