How much energy is required to change a 25 g ice cube from ice at -10°C to steam

Question

How much energy is required to change a 25 g ice cube from ice at -10°C to steam at 125°C? Answer in J.

I have m=.025kg
specific heat of ice = 2090 J/kgC
change in temperature from beyond freezing to freezing = 10 C
latent heat of fusion = 3.33x10^5 J/kg
specific heat of water = 4186 J/kgC
change in temperature from freezing to boiling = 100 C
latent heat of vaporization = 2.26x10^6
specific heat of steam = 2010 J/kgC
change in temperature from boiling to vaporization = 25 C

I got 66808.4 J, and this answer is wrong. What am I doing wrong?

Explanation / Answer

We can break this up into 5 stages. Stage 1 = Heating the ice to 0 degrees. Stage 2 = Melting the ice. Stage 3 = Heating the water to 100 degrees. Stage 4 = Turning the water into steam. Stage 5 = Heating the steam to 125 degrees. The energy required is as follows: Stage 1 = (.025kg)(2090 J/kgC)(10C) = 522.5J Stage 2 = (.025kg)(3.33 x 10^5 J/kg) = 8325J Stage 3 = (.025kg)(4186 JkgC)(100C) = 10465J Stage 4 = (.025kg)(2.26 x 10^6 J/kg) = 56500J Stage 5 = (.025kg)(2010 J/kgC)(25C) = 1256.25J Adding all these up gives a total energy of 77068.75J

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