Posted by **Claudia** on Sunday, February 1, 2009 at 1:54pm.

Given:

cp,ice = 2090 J/kg◦C

cp,water = 4186 J/kg ·◦C

cp,steam = 2010 J/kg ◦C

Lf = 3.33 × 105 J/kg

Lv = 2.26 × 106 J/kg

How much energy is required to change a

37 g ice cube from ice at −10◦C to steam at

116◦C? Answer in units of J.

ok.. so this is how i worked it out

37g(2090)(10C)=773300J

37g(3.33*10^5)=12321000J

37g(4186)(100)=15488200J

37g(2.26*10^6)=83620000J

37g(2010)(116-100)=1189920J

i added all of them together and i got

113392420J.

is this right?

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