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An 18.0 g piece of an unidentified metal was heated from 21.5 °C to 89.0 °C. If 789.75 J of heat energy was absorbed by the metal in the heating process, what was the identity
of the metal?

Respuesta :

Answer: The metal is Calcium.

Explanation:

To calculate the specific heat of substance during the reaction.

[tex]q=m\times c\times \Delta T[/tex]

where,

q = heat absorbed = 789.75 J

c = specific heat of metal = ?

m = mass of substance = 18.0 g

[tex]\Delta T_f[/tex] = final temperature  - initial temperature  = [tex](89.0-21.5)^0C=67.5^0C[/tex]

Now put all the given values in the above formula, we get:

[tex]789.75J=18.0g\times c\times 67.5^0C[/tex]

[tex]c=0.65J/g^0C[/tex]

As specific heat is characteristic of each metal and thus the metal is calcium which has  specific heat of [tex]0.65J/g^0C[/tex]

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