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

[physics1234]

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.

 

[OlderDan]

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).

 

[kyle8921]

I would recommend using the specific heat capacity (c) for each phase change instead of the latent heat (L). The specific heat capacity takes into account the energy required to change the temperature of a substance, while the latent heat only accounts for the energy required to change the phase of a substance.

The correct equation to use in this scenario would be Q = mcΔT, where Q is the energy added, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the first phase change from ice to water, the equation would be Q = (0.055 kg)(2.09 kJ/kg°C)(100°C - 0°C) = 11.495 kJ.

For the second phase change from water to steam, the equation would be Q = (0.04 kg)(4.18 kJ/kg°C)(100°C - 100°C) = 0 kJ.

Therefore, the total energy added to accomplish the transformation would be 11.495 kJ.