0.020 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 x 10^6 J/kg. specific latent heat of ice = 3.4 x 10^5 J/kg. specific heat capacity of water = 4.2 x 10^3 J/kg/K).
0.020 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 x 10^6 J/kg. specific latent heat of ice = 3.4 x 10^5 J/kg. specific heat capacity of water = 4.2 x 10^3 J/kg/K).
Please help!
What have you tried?
The question is a bit tricky regarding interpreting “how much water is now in the container”.
What you need is: Energy lost by the steam = Energy gained by the ice.
The question is a bit tricky regarding interpreting “how much water is now in the container”.
What you need is: Energy lost by the steam = Energy gained by the ice.
Thank you! But isn't energy also gained or lost by the water that is already there? (lost by ice, gained by steam, so whether it gains or loses energy in total depends on how much ice is VS how much water there is).
Thank you! But isn't energy also gained or lost by the water that is already there? (lost by ice, gained by steam, so whether it gains or loses energy in total depends on how much ice is VS how much water there is).
Or am I wrong? Thank you
The (liquid) water at 0°C is a distraction.
The interpretation that I have in mind is that the final equilibrium is 0°C.
You may want to show your work so that we can discuss it from there.
0.020 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 x 10^6 J/kg. specific latent heat of ice = 3.4 x 10^5 J/kg. specific heat capacity of water = 4.2 x 10^3 J/kg/K).