Question

If 1.25 g of water is heated from 20.0 degrees Celsius to 45.0 degrees Celsius and has a specific

heat of 4.18 J/gC, how much heat energy does it contain?

Answer 1

Answer: Heat energy contained by water is 130.625 J.

Step-by-step explanation:

Given: Mass = 1.25 g

Specific heat =
`4.18 J/g^(o)C`

Initial temperature =
`20.0^(o)C`

Final temperature =
`45.0^(o)C`

Formula used to calculate heat energy is as follows.

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

where,

q = heat energy

m = mass of substance

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 1.25 g * 4.18 J/g^(o)C * (45.0 - 20.0)^(o)C\\= 130.625 J`

Thus, we can conclude that heat energy contained by water is 130.625 J.