PLEASE HELP ME WITH THIS ONE QUESTION What is the final temperature when 625 grams of water at 75.0° C loses 7.96 x 10^4 J? (cwater = 4186 J/kg°C)

44.57°C

Explanation:

The following can be deduced from the question:

Specific heat of water = 4.186 J/kg

From the question, we can infer that 625 × 4.186 joules of heat will be lost when there's a 1°C drop of water.

We then calculate the amount if degrees that it'll take to cool for 7.96 x 10⁴J. This will be:

= 7.96 × 10⁴ /(625 × 4.186)

= 79600/(625 x 4.186)

= 79600/2616.25

= 30.43°C

The final temperature will then be:

= 75.0°C - 30.43°C

= 44.57°C

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