Physics Problem

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i had a problem on a physics test (high school physics) today that is absoulutely killing me. i got an answer relatively easily only to come out of the test hearing other students got similiar but different answers. i couldnt find an answer through the research powers of google so i thought i would come here to ask it. sorry if i insult anyones intelligence by presenting this problem Very Happy

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How many grams of steam would it take to melt 500g of ice (0degrees C) to 0 degree C water?

many of the answers people got were like: 74, 62.5, 55.5, 400

thanks for the help!
-Tommy

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Qin=Qout

Heat of Fusion of water @ 0 degC=334Kj/kg

Qin=500g*1kg/100g*334kJ/kg=167kJ to melt the ice @ 0degC

Heat of vaporization of steam @ 100 degC= 2260 kJ/kg

Qout=167kJ=ms kg * 2260kJ/kg

So ms (kg) = 167 kJ/2260 kJ/kg=0.074 kg or 74 g 100 degC steam to melt the ice

Now the melted ice and the condensed steam will mix and come to a final equilibrium temp. This temperature used the heat capacity of water (4.7kJ/kg/degC)

500(T-0)degC=74(100-T)degC
or
574T=7400 degC
so T=12.9 degC

Rap

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Huh?

1) If you're starting with water ice at 0 degrees C (=273 deg K) then it is ice VI, which is not your typical ice...it implies high pressurized conditions of at least 600 MPa (6 billion pascals), which is 60,000 times more pressure than atmospheric pressure at sea level.

2) If you're changing ice at 0 deg to water at 0 deg, this phase change is going to be due to relieving pressure...and adding steam to a closed system would be INCREASING pressure.

3) You never said how hot the steam was either.

This question just doesn't make any sense...

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thank you raprap, your explanation was very helpful.

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As Stuh said, unless we know the temperature of the steam (perhaps 100 degrees C?), the question has no answer.

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This is a HS physics question. Superheated and saturated high pressure steam is beyond the scope. Here it is best to assume near room conditions.

Rap

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stuh505 wrote:

Huh?

1) If you're starting with water ice at 0 degrees C (=273 deg K) then it is ice VI, which is not your typical ice...it implies high pressurized conditions of at least 600 MPa (6 billion pascals), which is 60,000 times more pressure than atmospheric pressure at sea level.

2) If you're changing ice at 0 deg to water at 0 deg, this phase change is going to be due to relieving pressure...and adding steam to a closed system would be INCREASING pressure.

3...

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If this is a school physics question none of the complexity above is applicable to the pedagogical exercise intended.

Simplistically
Heat lost by Steam at 100 changing to Water + Heat lost by water from steam cooling to zero = Heat gained by 500g ice changing to water at zero.

M x 2260 + M x100 x 4.18 = O.5 x 334

Whence M =appx 0.062kg = 62g.

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fresco wrote:

If this is a school physics question none of the complexity above is applicable to the pedagogical exercise intended.

Simplistically
Heat lost by Steam at 100 changing to Water + Heat lost by water from steam cooling to zero = Heat gained by 500g ice changing to water at zero.

M x 2260 + M x100 x 4.18 = O.5 x 334

Whence M =appx 0.062kg = 62g.

Good refinement of the problem

Rap

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