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A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C^−1 , is put into a calorimeter (see sketch at right) that contains 300.0g of water. The aluminum sample starts off at 94.5°C and the temperature of the water starts off at 21.0°C .When the temperature of the water stops changing it's 23.8°C .The pressure remains constant at 1atm .Calculate the mass of the aluminum sample.

Respuesta :

Answer: The mass of aluminium sample is 55.4 gram

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 aluminium = ?

[tex]m_2[/tex] = mass of water = 300.0 g

[tex]T_{final}[/tex] = final temperature = [tex]23.8^0C[/tex]

[tex]T_1[/tex] = temperature of aluminium = [tex]94.5^oC[/tex]

[tex]T_2[/tex] = temperature of water = [tex]21.0^oC[/tex]

[tex]c_1[/tex] = specific heat of aluminium = [tex]0.897J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of water =  [tex]4.184J/g^0C[/tex]

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]-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)][/tex]

[tex]m_1=55.4g[/tex]

Therefore, the mass of aluminium sample is 55.4 gram