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maks197457 [2]
3 years ago
6

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.
Chemistry
1 answer:
sineoko [7]3 years ago
8 0

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

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\times c\times \Delta T

Q = Heat absorbed=6832 Joules

m= mass of water = 5.9mol\times 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\times 4.184\times (T_f-18.6)

T_f=34.0^0C

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

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