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You are performing an experiment where you place 6.7797 g of Al metal at 205.24 equationC into a coffee cup calorimeter that contains 87.4 g of water at 25.37 equationC. The final temperature of the water in the coffee cup is 28.67 equationC. Remember, the heat gained by the water is equal to the heat lost by the Al. equation What is the specific heat (C) of the Al in J/g°C? Cwater= 4.184 J/g°C. Do not include units. If you need to express your answer as an exponential number, use this template: 1445 should be typed as 1.445e+003

Respuesta :

Answer: The specific heat of the Al is [tex]1.008J/g^0C[/tex].

Explanation:

[tex]Q_{absorbed}=Q_{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\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]      

where,

[tex]m_1[/tex] = mass of water = 87.4 g

[tex]m_2[/tex] = mass of Al metal = 6.7797 g

[tex]T_{final}[/tex] = final temperature = [tex]28.67^0C[/tex]

[tex]T_1[/tex] = temperature of water = [tex]25.37^oC[/tex]

[tex]T_2[/tex] = temperature of Al metal = [tex]205.24^oC[/tex]

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

[tex]c_2[/tex] = specific heat of Al metal = ?

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

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

[tex]87.4\times 4.184\times (28.67-25.37)^0C=-[6.7797\times c_2\times (28.67-205.24)][/tex]

[tex]c_2=1.008J/g^0C[/tex]

Therefore, the specific heat of the Al is [tex]1.008J/g^0C[/tex].

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