Question

How much heat is added if .617g of water is increased in temperature by .241 degrees C?

Answer 1

0.622 J

Explanation:

In order to solve this problem, you need the formula:
`q=mC_sDeltaT`

Assuming water is in its liquid state, the specific heat capacity or `C_s` would be 4.18 `J/(g* degrees Celsius)`

`DeltaT` in this case would be 0.241 degrees Celsius since this represents the change in temperature.

`m` would be 0.617 grams.

Plug in all the values and you would get 0.622 joules!
Hope this helps!