How much heat is required to turn ice into water vapour?

This is the question:

"How much heat is required to turn 1.5 kg of water ice at a temperature of −25°C into 1.5 kg of water vapour at 150°C? (to 2 s.f and in MJ)

[cice = 2090 J kg−1 K−1, cwater = 4190 J kg−1 K−1, cwater-vapour = 2010 J kg−1 K−1, Lvap-water = 2256×103 J kg−1, Lfus-water = 333×103 J kg−1]"

So I used Q=mL twice - one for vap-water and one for fus-water.

I used Q=mc/_\T three times, each for each variation of specific heat capacity (c).

I kept the change in temperature constant - from -25C to 150C, so a change of 175 degrees celsius total.

I used Q1+Q2+Q3... for the last step and so added the results of all of these together which came to 6059625J, which to MJ is 6.1MJ.

It's definitely within the MJ range, so I thought it'd be right but it's not.
I wonder where I went wrong here, (Shake) any idea?

Jun 2016
You seem to have the essential ideas correct,
Although I am a bit confused by your statement about keeping the temperature change constant.

You should have 25C with Ice, 100C with water, 50C with vapour
plus the two phase changes (ice to water, water to vapour).
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I see!!

That you very much. I got a way bigger number (4400MJ!) but I believe it is correct. I will remember to take into account that temperature must change between these calculations when needed. I also assume I will have to remember the temps phase change occurs at, lol. Totally never would have thought of and learn ;P

Thanks agin!