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.

Step-by-step explanation:

To calculate the specific heat of substance during the reaction.

`q=m* c* \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* c* 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`