Question

At 1 atm, how much energy is required to heat 51.0 g of H2O(s) at –22.0 °C to H2O(g) at 131.0 °C? Helpful constants can be found here.

Answer 1

step 1 ) convert ice from -22⁰C to 0⁰C

q1 = mC_iceT

syep2) convert ice to water at 0⁰C

q2 =mH_fus

step3) convert water at 0⁰C to 100⁰C

q3 = mC_water T

step 4 ) convert water to vapour at 100⁰C

q4 = mH_vap

step5) convert vapour at 100⁰C to 131⁰C

total heat required q = q1 + q2 + q3 + q4 + q5

q = mC_iceT + mH_fus   + mC_waterT + mH_vap   + mC_vapT

q = 51 x 2.05 x 22 + 51 x 333.55 + 51 x 4.186 x 100 + 51 x 2258.3 + 51 x 1.996 x 31

q = 2300.1 + 17011.05 + 21348.6 + 115173.3 + 3155.676

q = 158988.726 J

q = 158.988 KJ