How much heat, in joules and in calories, must be removed from 3.61 mol of water to lower its temperature from 23.7 to 11.5°C? Include the sign in your answer.
joules
calories
q = msT
mass of water = moles * molarmass
= 3.61*18
= 64.98 grams
q = - 64.98*4.184*(23.7-11.5)
= - 3316.89 J
= - 3.31689 KJ
= - 792.75 calories
= - 0.79275 Kcalories
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