calculate the total energy transferred when cooling a 50 gram sample of H2O from

Question

calculate the total energy transferred when cooling a 50 gram sample of H2O from 25 celcius to -25 celcius. Asumme no energy is lost to the surroundings

C(g)= 1.996 j/g.Celsius C(l)= 4.184 J/g.Celsius C(s)= 2.108 J/g.Celcius Hvap= 40.68 KJ/mol Hfus= 6.02KJ/mol Melting pt: 0 celcius Boiling pt= 100 celcius MW H20= 18.02 g/mol

I'm reposting this question so someone else can be clear from where all the values come from please. Imagine you are helping someone who doesnt get it all fast. Thanks in advance!!

Explanation / Answer

Answer:

Total energy transfer = mass of water (g)× 4.2 × T (o C)

Here mass of water = 50 g

Temperature = 25 o C

Therefore, total energy transfer = 50×4.5× 25 = 5250 J

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