Question

What is the specific heat of a 22.8 g sample of metal that absorbs 1,450 Joules of heat from 21.8 degrees Celsius to 75.0 degrees Celsius?

Answer 1

Answer: The specific heat of given metal is
`1.195 J/^(o)C`.

Step-by-step explanation:

Given: Mass of sample = 22.8 g

Heat energy = 1450 J

Initial temperature =
`21.8^(o)C`

Final temperature =
`75^(o)C`

Formula used to calculate the specific heat is as follows.

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

where,

q = heat energy

m = mass of substance or sample

C = specific heat

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\1450 J = 22.8 g * C * (75 - 21.8)^(o)C\\C = (1450 J)/(22.8 * 53.2^(o)C)\\= 1.195 J/^(o)C`

Thus, we can conclude that the specific heat of given metal is
`1.195 J/^(o)C`.