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

Question

3 years ago

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of the water will be _____________________0C. Record your answer to 1 decimal place.

1 answer:

sineoko [7] · Answer 1 · 2022-03-08T13:58:02+03:00

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 $34.0^0C$