ice to water to 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).