Homework Help: Physics

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?

Physics - drwls, Sunday, February 1, 2009 at 2:26pm
113,392,420 J looks good to me.

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