Question

1) A sample of steam with a mass of 0.505 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.25 g of water at 3.0∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)

Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture? In celcius!

2) Nanomaterials are materials that have dimensions on the 1- to 100-nm scale. Carbon, metals, and semiconductors exhibit distinct properties on the nanoscale that differ from their properties under other conditions.

Gold has a face-centered cubic arrangement with a unit cell edge length of 4.08 Å . How many moles of gold fit in a gold nanoparticle sheet with a length of 87.1 nm , a width of 21.7 nm , and a thickness of 19.9 nm ?

Answer 1

msteam = 0.505*2260 = 1141.3 due to condensation

Qc = m*Cp(Tf-Ti) = 4.25*4.184*(Tf-3) - 1141.3

Qh = m*Cp*(Tf-Ti) = 0.505*4.184*(Tf-100)

qc = -qh

4.25*4.184*(Tf-3) - 1141.3 = -0.505*4.184*(Tf-100)

17.782Tf- 3*17.782 - 1141.3 = -2.11292Tf + 2.11292*100

17.782Tf - 1087.954 = -2.11292Tff + 211.292

(17.782 + 2.11292)Tf = 211.292+1087.954

Tf = (211.292+1087.954) /((17.782 + 2.11292) = 65.3054 °C

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