Question

What is the number of joules of heat energy released when 20 grams of water is cooled from 293 K to 283 K?

Answer 1

836 J

Explanation:

Use the formula `q = mCΔT`

q = heat absorbed or released, in joules (J)
m = mass
C = specific heat capacity
ΔT = change in temperature

Plug known values into the formula.

The specific heat capacity of water is `4.18 J/g * K`.

`q = 20(4.18)(293 - 283)`
`q = 20(4.18)(10)`
`q = 836`

836 joules of heat energy are released.