Question

A sample of steam with a mass of 0.501 g at a temperature of 100°C condenses into an insulated container holding 4.45 g of water at 4.0°C. (For water, ΔH°vap = 40.7 kJ/mol and Cwater = 4.18 J/(g⋅°C).) What is the final temperature?

          A sample of steam with a mass of 0.501 g at a temperature of 100°C condenses into an insulated container holding 4.45 g of water at 4.0°C. (For water, ΔH°vap = 40.7 kJ/mol and Cwater = 4.18 J/(g⋅°C).) What is the final temperature?
        
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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A sample of steam with a mass of 0.501 g at a temperature of 100°C condenses into an insulated container holding 4.45 g of water at 4.0°C. (For water, ΔH°vap = 40.7 kJ/mol and Cwater = 4.18 J/(g⋅°C).) What is the final temperature?
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Transcript

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00:01 Of steam and the mass of the steam is 0 .501 grams.
00:11 My initial temperature is given as 100 degrees c.
00:17 It condenses and it has it into a cup, an insulated cup, that has the mass of water of 4 .45 grams and the initial temperature of the water is 4 .0.
00:35 Degrees c.
00:38 Find the final temp.
00:49 Okay, so we know that the heat of vaporization for water is given as 40 .7 kilojoules per mole and my c of water is given as 4 .18 j over g degrees c.
01:16 So we are going to figure out how much energy is going to be required to vaporize the steam or to condense the steam.
01:35 Sorry about that.
01:43 And that'll be, the energy required to do that will be 0 .501 grams divided by 18 .02 grams per mole.
01:55 This will give me my moles of water times 40 .7 kilojoules per mole times 1 ,000 joules per kilojoule.
02:12 And this will give me 0 .501 times 40 .7 times 1 ,000 divided by 18 .02.
02:26 And that will be 1131 .6 joules to condense steam to 100 degrees c.
02:46 So now let's figure out how much that will raise my temperature of water...
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