How Much Energy is Needed to Transform Ice into Water and Steam?

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SUMMARY

The discussion focuses on calculating the energy required to transform a 55 g ice cube at 0°C into water and steam. The correct approach involves using the latent heat of fusion and vaporization, along with the specific heat capacity of water. The calculations should include 55 g for melting, heating the resulting water from 0°C to 100°C, and vaporizing 15 g of steam. The total energy added is 52215 J, which was initially miscalculated due to incorrect application of the formulas.

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A 55 g ice cube at 0°C is heated until 40 g has become water at 100°C and 15 g has become steam at 100°C. How much energy was added to accomplish the transformation?

I tried using Q=mL for each change and adding them together.

Q=.055(3.33x10^5)
Q=18315

then

Q=.015(2.26x10^6)
Q=33900

18315+33900= 52215 which is not the correct answer.
 
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You have 55g melting (latent heat of fusion) and 55g warming from 0° C to 100° C and then 15g vaporizing (latent heat of vaporization).
 

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