Solving 0.02 kg Ice and 0.10 kg Water at 0°C to Melted State

[risingabove]

Homework Statement

0.02 kg of ice and 0.10 kg of water at 0 degrees Celsius are in a container. steam at 100 degrees Celsius is passed in until all the ice is just melted. How much water is now in the container?

Specific latent heat of steam = 2.3 * 10^6 J/kg
Specific latent heat of ice = 3.4 * 10^5 J/kg
Specific heat capacity of water = 4.2 * 10^3 J/kg/K

Homework Equations

latent heat Q = mL
Specific Heat Capacity Q = mCT

Where Q = Heat required
m = mass of substance
C = Specific heat Capacity
T = change in Temperature

The Attempt at a Solution

Using latent heat equation, the heat required for a phase change from ice to water

Q = 0.02 (3.4 x 10^5)
= 6800 J

Using Specific heat capacity equation, the heat require for water

Q = 0.10(4.2 x 10^3)100
= 42,000 J

Total heat required

6800 + 42000 = 48,800 J

plugging in 48,800 into latent heat equation for steam

48,800 = m (2.3 x 10^6)
m = 0.0212 kg

will this answer be the mass of the water in the container?
or am i missing steps or is completely wrong with my workings?

 

[SteamKing]

Units, units, units! Always includes units for all the given quantities!

 

[BvU]

I understand you want to use some heat to melt the ice. But what is it you want to do with the water ?

 

[risingabove]

I am trying to find out how much water now in the container after the steam as passed in...the answer suppose to be 0.1225 kg , I am not understanding where i must have gone wrong... or if i missed out a step.

 

[BvU]

Repeat: it looks as if you want to heat the water to 100 degrees. Why ?

 

[risingabove]

Well i was thinking in lines of finding out the heat required to raise the temperature of water in the container to 100 degrees celsius. since that will be the new temperature in the container when steam passes through ... finding that heat required for the change of water temperature, i added it to the heat required for the phase change from ice to water to give me total amount of heat in the system...which i then used in the equation for latent heat of steam...

 

[BvU]

The exercise asks for the amount of heat to just melt the ice (6800 J as you calculated). Water and ice are at equilibrium at 0 C, so if the last bit of ice melts, the temperature is still 0 C ! In other words: the whole thing is played out at 0 C. Steam condenses at 100 C, so it sure condenses at 0 C. Heat given off is latent heat plus ... ?

Pity you already know the answer; something of a spoiler.

It is remarkable how much bigger the latent heat at phase change liq -> vapour is than at solid -> liquid ! A factor of 7. So even if you miss the plus ... you expect not to need more than 3 grams of steam to melt 20 grams of ice. 100 + 20 + 3 = 123 grams, quite close to the right answer of 122.5 grams...

 

[risingabove]

OMGGGGGG... i can't believe i missed something soo simple...arggggghhh ...thanks tho...

 

[BvU]

Sometimes lines of thought are so powerful they become blocking. If you can learn to challenge your own ideas equally well as you are able to challenge those of others, you gain something very precious. Lateral thinking is too modest a term for it. Wisdom comes with age, but the earlier you start the better .