Question

At 1 atm, how much energy is required to heat 37.0 g H2O(s) at −20.0 ∘C to H2O(g) at 133.0 ∘C?

Answer 1

1.Heat required to raise temperature of 37 g H₂O from -20⁰C to 0⁰C :

q1=mc(T₂-T₁)=37*2.09J/g*20=1546.6J

2.Heat required to melt 37g ice at 0⁰C =q2=mL_fus=37*334J/g=12358J

3.Heat required to raise temperature of 37 g H₂O from 0⁰C to 100⁰C:

q3=mc(T₂-T₁)=37*4.184J/g*100=15480.8J

4.Heat required to convert 37 g liquid to gas at 100⁰C:

q4=mH_vap=37*2260J/g=83620J

5.Heat required to raise temperature of 37 g steam from 100⁰C to 133⁰C=

q5=mc(T₂-T₁)=37*1.9J/g*(133-100)=37*1.9*33=2319.9J

for total amount of heat add q1+q2+q3+q4+q5=115325.3J=115.3252KJ