At 1 atm, how much energy is required to heat 71.0 g of H2O(s) at –22.0 °C to H2O(g) at 125.0 °C?

Question

I have no idea of what I am doing here.

Answer 1

Hi Denise,

 

If you have Chemistry textbook, go into the section "Calorimetry" for this specific problem.

 

 q = ms Δ t     (equation for calculating the heat change)

 

q = (71.0g) (4.181 J/g C) (125 C - -22 C)

 

q = (71.0g) (4.184 J/g C) (125 C + 22 C)

 

q = (71.0g) (4.184 J/g C) (147 C)

 

q = 43668.408 J

 

q = 43.7 kJ

 

I hope you find it and get it.