Question

Calculate what the specific heat capacity of an unknown metal is if 394)

of heat were released when 15.0 grams of the substance at 96.0°C was
cooled to 28.0°C?
Formula Reminders
q = mCAT
q = mHvap
q = mHfus
1. 4.01 J/gºC
2. 5910 J/g°C
3. 0.387 J/g°C

Answer 1

Answer: The specific heat capacity of an unknown metal is
`0.387 J/g^(o)C`.

Step-by-step explanation:

Given: Heat energy = 394 J

Mass = 15 g

Initial temperature =
`96^(o)C`

Final temperature =
`28^(o)C`

Formula used is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = heat energy

m = mass of substance

C = specific heat capacity

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\394 J = 15 g * C * (28 - 96)^(o)C\\C = (394 J)/(15 g * (68^(o)C))\\= (394 J)/(1020 g^(o)C)\\= 0.387 J/g^(o)C`

Thus, we can conclude that the specific heat capacity of an unknown metal is
`0.386 J/g^(o)C`.