Answer to Question #103340 in General Chemistry for mya

Question

At 1 atm, how much energy is required to heat 85.0g H2O(s) at −20.0∘C to H2O(g) at 153.0 C?

Answer 1

We will calculate the following steps:

1. For heating ice from -20 ° C to 0 ° C:

(ΔT = 20 ° C):

Q = mcΔT: m - mass; c - specific heat: ΔT- temp. difference; Q - heat is absorbed or released.

Q = 85.0 g x 2.1 J / g / ° C x 20 ° C = 3 570 J= 3.570 kJ

2. Melt ice at 0 ° C to water at 0 ° C:

Q = mK: m is the melting mass.

= 85.0 g x 334 J / g = 28 390 J = 28.390 kJ

3. It is allocated for heating water from 0 ° C to boiling at 100 ° C:

Q = mcΔT: m - mass; c - specific heat: ΔT - temp. difference; Q -heat is absorbed or

= 85.0 g x 4.184 J / g / ° C x 100 ° C ΔT = 35 564 J = 35.564 kJ

4. To evaporate water into steam at 100 ° C:

Q = mK

= 85.0 g x 2260 kJ / g = 192.100 kJ

5.It is allocated for heating steam from 100 ° C to 156 ° C:

Q = mcΔT: m = mass; c = specific heat: ΔT = temp. difference; Q = heat is absorbed or

= 85.0 g x 2.01 J / g / ° C x 53 ° C ΔT = 9 055.05 = 9.055,05 kJ

Q = 3.570 kJ + 28.390 kJ + 35.564 kJ + 192.100 kJ + 9.055,05 kJ=268.679,05 kJ