Question

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?

Answer 1

Answer: The metal is Calcium.

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times \Delta T$

where,

q = heat absorbed = 789.75 J

c = specific heat of metal = ?

m = mass of substance = 18.0 g

$\Delta T_f$ = final temperature  - initial temperature  = $(89.0-21.5)^0C=67.5^0C$

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

$789.75J=18.0g\times c\times 67.5^0C$

$c=0.65J/g^0C$

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