Question

6832 J of heat energy is applied to 5.9 mol of water. If the original temperature of the water was 18.60C, the final temperature of the water will be _____________________0C. Record your answer to 1 decimal place.

Answer 1

Answer: The final temperature of the water will be
`34.0^0C`

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`Q=m* c* \Delta T`

Q = Heat absorbed=
`6832` Joules

m= mass of water =
`5.9mol* 18g/mol=106.2g`

c = specific heat capacity =
`4.184J/g^0C`

Initial temperature of the water =
`T_i` =
`18.6^0C`

Final temperature of the water =
`T_f` = ?

Putting in the values, we get:

`6832=106.2* 4.184* (T_f-18.6)`

`T_f=34.0^0C`

The final temperature of the water will be
`34.0^0C`