Question

How much heat (in kj) is needed to convert 866g of ice at -15*C to steam at 146*C? (Note: The specific heat of ice is 2.03J/g*C and the specific heat of steam is 1.99J/g*C. Delta fus for H20 is 6.01 kj/mol and Hvap is 40.79kj/mol)

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

How much heat (in kj) is needed to convert 866 g of ice at -10C to steam at 126 C?

(The specific Heat of Ice, water, and steam are 2.03 j/g C,4.184 j/g C, and 1.99 j/g C respectively.)

The heat of fusion of water is 6.01 kj/mol, the heatvaporization is 40.79 kj/mol.

Bring ice to 0 C
q = mass*cp*delT

q = 866 g *(2.03 J/g/C) *(0 - (-10))C
q = 17579.8 J

Melt ice

q = delH*mass
q = 6.01 kJ/mol *866 g*(1000 J/1kJ)*( 1mol /18g)
q = 289147.778 J

Bring water to 100 C
q = 866 g *4.184 J/g-C *(100 - 0) C
q = 362334.4 J

Vaporize water
q = 866 g*(1 mol/18 g) *(40.79kJ/mol )*(1000 J/1kJ)
q = 1962452.22 J

bring steam to 126 C

q = 866 g *(1.99 J/g-C)*(126 - 100 )C
q = 448378.84 J

Sum of q = 17579.8 + 289147.778 + 362334.4 + 1962452.22 +448378.84
Q = 3079893.04 J ~ 3.08 e3 kJ