How much heat (in kj) is evolved in converting 1.00 mol of steam at 160.0∘c to ice at -50.0∘c? the heat capacity of steam is 2.01 j/g⋅∘c and of ice is 2.09 j/g⋅∘c.

58.266kJ

Explanation:

Let's bring out the information given in the question;

Number of mol = 1

Temperature of Steam = 160∘C

Temperature of ice = -50∘C

Heat capacity of steam = 2.01J/g⋅∘C

Heat capacity of ice = 2.09J/g⋅∘C

Mass = Number of moles * molar mass = 1  * 18g/mol = 18g

It's important to be able to understand the phase changes that are happening and associate the energies with them.

The following changes and their energy associated with them are calculated below;

Steam would have to cool down to 100∘C

Formular for this energy change is given as; H = MC∆T

M = 18g

C = 2.01J/g⋅∘C

∆T = T2 - T1 = 160 - 100 = 60∘C

H1 = 18 * 2.01 * 60 = 2170.8J

Steam would change to liquid at 100

H = mL

m = 18g

L (Latent heat of vaporization) = 2259.3 J/g

H2 = 18 * 2259.3 = 40667.4J

Liquid water would then cool to 0 from 100

Formular for this energy change is given as; H = MC∆T

M = 18g

C = 4.186J/g⋅∘C

∆T = T2 - T1 = 100 - 0 = 100J/∘C

H3 = 18 * 4.186 * 100 = 7534.8J

Liquid water would change phase to ice at 0

H = mL

m = 18g

L = 334J/g

H4 = 18 * 334 = 6012J

Ice would then have to cool to -50

Formular for this energy change is given as; H = MC∆T

M = 18g

C = 2.01J/g⋅∘C

∆T = 0 - (-50) = 0 - 50 = 50

H5 = 18 * 2.09 * 50 = 1881J

Total Heat Evolved = H1 + H2 + H3 + H4 + H5 = 58266J = 58.266kJ