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A sample of steam with a mass of 0.530 g at a temperature of 100 ∘C condenses into an insulated container holding 4.20 g of water at 6.0 ∘C. (For water, ΔH∘vap=40.7 kJ/mol and Cwater=4.18 J/(g⋅∘C).)

Answer :

hopemichael95

Answer:4.66C

Explanation:

q1 heat gained by water = mcdeltaT

massH2O x specific heat water x (Tfinal-Tinitial). T f is the final Temperature and Ti is the initial Temperature.

q1 = 0.530g x 4.184 J/g*C x (Tfinal - 100.0) = 0.022

q2 = heat lost by steam in converting steam (g) at 100.0 C to liquid at 100.0 C = mass x delta H.

q2 = 4.20g x 40.7 (Tfinal - 100.0) = 1.71

q3 = heat lost with 4.20 g (from the steam conversion) water moving from 6.0 C to final T.

q3 = 4.20g x 4.184 J/g C x (t final - 6.0) =2.93

Ti is 6.0

t final= q1 + q2 + q3

0.022 + 1.71 + 2.93 = 4.66 C

Tfinal= 4.66C

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