If 20 grams of lead vapor at 1750 degrees celsius is added to 20 grams of ice at

ID: 2252012 • Letter: I

Question

If 20 grams of lead vapor at 1750 degrees celsius is added to 20 grams of ice at -100 degrees celsius

A) What is the final temperature of the mixture after it has reached thermal equilibrium?

B) Determine how much of the lead and H2O will be in the solid/liquid/gaseous state?

Constants:

Lead boils at 1750 degrees celsius

Latent heat of vaporization of lead is 8.58x10^5 J/Kg

Lead melts at 328 degrees celsius

Latent heat of Fusion water- 3.33x10^5 J/Kg

Latent heat of Vaporization- 22.6x10^5 J/Kg

Specific Heats:

Lead liquid-450 J/Kg

Ice- 2090 J/Kg

Water- 4190 J/Kg

Steam- 1996 J/Kg

Heat released

(Lead gas at 1750 to lead liquid at 1750) = 0.02*8.58*10^5 = 17.16 kJ

(Lead liquid at 1750 to lead liquid at 328) = 0.02*450*(1750-328) = 12.798 kJ

(Lead liguid at 328 to lead solid at 328) = 0.02*23020 = 0.460 kJ

(Lead solid at 328 to Lead solid at 100) = 0.02*128*(328-100) = 0.584 kJ

Total heat released = 31.002 kJ

Heat absorbed

(ice at -100 to ice at 0) = 0.02*100*2090 = 4.18 kJ

(ice at 0 to water at 0) = 0.02*3.33*10^5 = 6.66 kJ

(water at 0 to water at 100) = 0.02*4190*100 = 8.38 kJ

(water at 100 to steam at 100) = 0.02*22.6*10^5 = 45.2 kJ

Total heat absorbed = 64.42 kJ

In this state, both the lead(s) and water(g) are at 100 C. But since heat is not balanced equillibrium is not achieved in terms of state.

But the final temp will be 100 C

Heat Difference = 64.42 - 31.002 = 33.418 kJ

Now this much heat was less, so only some water will convert to steam or in other words, steam requiring this much heat will condense back.

m = Heat/2260 = 33.418/2260 = 0.01478 kg

So final Equillibrium Composition:

water (100 C) = 14.78 g

Steam(100 C) = 5.22 g

Lead (100C) = 20 g

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