Using specific heat stuff to find amount of heat(Q=mCT)

[Raziel2701]

Homework Statement

Beryllium has roughly one-half the specific heat of liquid water (H2O). Rank the quantities of energy input required to produce the following changes from the largest to the smallest. In your ranking, note any cases of equality. (Use only the symbols > or =, for example a>b=c)

(a) raising the temperature of 1 kg of H2O from 20°C to 26°C
(b) raising the temperature of 2 kg of H2O from 20°C to 23°C
(c) raising the temperature of 2 kg of H2O from 1°C to 4°C
(d) raising the temperature of 2 kg of beryllium from -1°C to 2°C
(e) raising the temperature of 2 kg of H2O from -1°C to 2°C

Homework Equations

Q=mC\Delta T

The Attempt at a Solution

For (a) I get 25116 J
For (b) I get 25116 J
For (c) I get 25116 J
For (d) I get 12558 J
For (e) I get 25116 J

So the answer I put in is a=b=c=e>d but it tells me it's incorrect. I found these answers simply by subtracting the final temperature from the initial, multiplying it by the mass and multiplying by the specific heat of water where relevant. For d I used 2093 as my specific heat, and for the rest I used 4186.

No clue what is wrong.

 

[mgb_phys]

(e) tricky, to go from water at -1 to 2, you have to heat ice from -1C to 0C and then melt it and then heat water from 0C-2C

Even without knowing the numeric value for the melting energy of ice you know that it takes energy

 

[Raziel2701]

Ah that's what I missed. Yeah that would make "e" the one to take the most energy since there would be a latent heat of fusion calculation in there.

Thanks!