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27.0g of an unknown metal at 100 degrees Celcius was dropped into a beaker containing 313g of water initially at 22.3 degrees Celcius. The final temperature of the mixture was 25.1 degrees. What is the specific heat capacity of the unknown metal?

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SvetkaChem

Answer:

[tex]1.81 J/g*^{o}C[/tex]

Explanation:

[tex]q = c_{p}*m*(T_{f} -T_{i})[/tex]

[tex]q _{m}= c_{p}_{m}*27.0g*(25.1^{o}-100^{o})[/tex]

[tex]\\\\q_{water} = c_{p}_{water*313 g*(25.1^{o} -22.3^{o})[/tex]

[tex]q_{m} = - q_{water}\\\\c_{p}_{m}*27.0g*(25.1^{o}-100^{o}) = - 4.18 J/g*^{o}C*313 g*(25.1^{o} -22.3^{o})[/tex]

[tex]c_{p}_{m}*2022.3 = 3663.352\\\\c_{pm}=1.81 J/g*^{o}C[/tex]

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