How much heat (in kj) is required to warm 11.0 g of ice, initially at -12.0 ∘c, to steam at 109.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘c?

Sensible heat (involving temperature change) is calculated as mc(deltaT), where c is the specific heat. Latent heat (involving phase change) is calculated as mH, where H is the latent heat of fusion or vaporization.
From ice at -12 to ice at 0: (11.0 g)(2.09 J/g-degC)(0 - (-12)degC) = 275.88 J
From ice at 0 to water at 0: (11.0 g)(80 J/g) = 880 J
From water at 0 to water at 100: (11.0 g)(4.184 J/g-degC)(100 - 0 degC) =4602.4 J
From water at 100 to steam at 100: (11.0 g)(540 J/g) = 5940 J
From steam at 100 to steam at 109: (11.0 g)(2.01 J/g-degC)(109 - 100 degC) = 198.99 J
Adding all the heats gives: 11897.27 J, which is equivalent to 11.89727 kJ.