Question 16 Equations until solid wator Consider heari iquid and th ga (ateam).

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Question 16 Equations until solid wator Consider heari iquid and th ga (ateam). (Figure The heat onergy rociotod with chango In tampamturo that charge in pharc given by tively, 1) Alton rocoso, cooling stoam Figure20 In each case, two types of transitione those involvi AT tamponaturo change no in phase Ishown by the diagonal ino sogmonts un where gis haatin jou mismas sin grams, s is specific heal in joules par gr celsius, J/(g. C) and AT is temperature dange in degraas Celsius. The heat enargy associated with change in the graph nd the6e at constant temperature with a change in phase (shownby phaeu a: constunt tormporatura ie givon he gr where yishsstiin jouea, mis ma sin grama, and All is ths entrepy injoules per gram. ical 2.00 J Spocito hast otice: AH Enthalpy o1 fusion Part A How much heatenergy. in kilooulea is requirodtoconvert 30.0 gcrice at -18.0 C to water at 25.0 C signilicant fig de the appropri Value Units Submit Hints My Answers Give up Ruview part Part B How long would it ke for 1.50 added Express yo wer to three significant figures de The apl Heat Value Units Submit Hunts My Answers Give up Review Part in Continue quilirium added U

Explanation / Answer

(A)

q1 = m * s * (t2-t1)ice = 39.0 * 2.09 * (0.000 + 18.0) = 1467.18 J

q2 = 39.0 * 334 = 713026 J

q3 = 39.0 * 4.18 * (25.0 - 0.000) = 4075.5

Therefore, total heat change = 1467.18 + 713026 + 4075.5 = 718568.68 J = 718. kJ

(B)

Enthalpy of vaporization of 1.50 mol of water = 1.50 * 18.0 * 225 = 6075 J

Time required to add 20.0 J of heat = 1 s

Then time required to add 6075 J of heat = 6075 / 20.0 = 303.75 s

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