At 1 atm, how much energy is required to heat 83.0 g of H2O(s) at –18.0 °C to H2O(g) at 129.0 °C?

Question

how do you work this problem?

Answer 1

To solve this question, use the q=mcΔt formula.  Here, m=83g, c=4.18 J/(g*K).  Δt is found by Tfinal-Tinitial, which is 129-(-18)=147 degrees C.  Since the conversion from C to K is only adding 273 to the C, the difference in degrees C is equivalent to degrees K. So:

 

q=(83g)*(4.18 J/g*K)*(147K)=51000.18J=51kJ