how many gram of ice at 0 c can be melted by the heat released when one gram of steam condense to water at 100 c
check my answer please and you deman i am sorry by accident i post it the previous question
solution
Qcold=-Qhot
MLf(ice)=-MlV(water)
Mice = -(-0.001*2256000)/(334000)
Mice melt=6.75 g
Yes, it is just the ratio of heat of vaporization to heat of fusion.
I get
2260 J/gram / 334 J/gram = 6.77 g
Heat out of steam= 20 {2260+4.2(100-T)}
Heat into water = 80 (T-20)
so
4(T-20) = 2260 + 4.2(100-T)
4 T - 80 = 2260 + 4200 - 4.2 T
8.2 T = 6540
T = 797 which is silly of course
What it means is that 80 grams of water at 20 degrees simply can not condense 20 grams of steam. The final temp is 100 deg C and you still have some stem left :)
There was another question in here that seems to have disappeared but my answer remained ?
To calculate the number of grams of ice that can be melted by the heat released when one gram of steam condenses to water at 100 degrees Celsius, you can use the equation:
Qcold = -Qhot
Where Qcold represents the heat absorbed (or released) by the cold substance, and Qhot represents the heat released (or absorbed) by the hot substance.
In this case, the hot substance is the steam condensing to water, and the cold substance is the ice melting.
The heat released when one gram of steam condenses to water is given by the latent heat of vaporization, which is 2256000 J/kg. Therefore, for one gram of steam, the heat released is 2256000 J.
The amount of ice that can be melted depends on the amount of heat absorbed by the ice, which is given by the equation:
Qcold = MLf(ice)
Where MLf(ice) represents the latent heat of fusion for ice, which is 334000 J/kg.
By rearranging the equation, we can solve for Mice (the mass of ice):
Mice = -Qcold / MLf(ice)
Plugging in the values:
Mice = -(2256000 J) / (334000 J/kg)
Mice = 6.75 g
Therefore, one gram of steam condensing to water at 100 degrees Celsius can melt 6.75 grams of ice at 0 degrees Celsius.