In a calorimetry experiment the water absorbs 273 J of heat (q), how much heat (q) is released by the metal if no heat is lost? Remember: -qreleased = qabsorbed ANS: kJ/mol

qreleased = -273J.

Explanation:

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In this case, according to this calorimetry question, and keeping in mind that the hot substance releases heat to the cold substance which absorbs it, according to:

-qreleased = qabsorbed

In such a wat, since the water absorbs 273 J of heat and no heat is lost, the amount of this measurement released by the metal is:

-qreleased = 273J  

qreleased = -273J

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93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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