Question

Steam at 100°C is added to ice at 0°C.

(a) Find the amount of ice melted and the final temperature when the mass of steam is 13.0 g and the mass of ice is 45.0 g.
 g
 °C

(b) Repeat this calculation, when the mass of steam as 1.30 g and the mass of ice is 45.0 g.
 g
 °C

Answer 1

Part (a)
First, think about melting all the ice at 0 degC to water, still at 0 degC.
Using the formula Q = mL, where m = 45 g = (45/ 1000) kg = 0.045 kg is the mass of ice and
L = 333 kJ/kg is the heat of fusion of water, we get
(1) .......... Q1 = ( 0.045 )( 333 ) = 14.985 kJ
Next, find out how much heat is needed to condense steam at 100 degC to water still at 0 degC.
Again, using the formula Q = mL, where m = 13 g = (13/ 1000) kg = 0.013kg is the mass of the available steam and L = 2256 kJ/kg is the heat of vaporization(condensation) of water, we get
(2) .......... Q2 = ( 0.013 )( 2256 ) = 29.328 kJ
for the amount of heat liberated in converting 13 g of steam at 100 degC to water still at 100 degC,
which is more than enough to melt all the ice present to water still at 0 degC.
It follows the remaining heat of
(3).......... Q2 - Q1 = 29.328 kJ - 14.985 kJ = 14.343kJ
can be used to further raise the temperature of the 45 g of water at 0 degC.
Using the formula Q = m*(SHCW)*(dT) where
Q = 14.343 kJ = 14.343e+03 J
m = 45g = 0.045 kg
SHCW = 4190 [ J / (kg*K) ] = 4190 [ J / (kg*degC) ]
(specific heat capacity of water)
we find that the change in temperature dT is
dT = Q / [ m*(SHCW) ] = 14.343e+03 / [ ( 0.045 ) ( 4190 ) ] = 76.07 degC

At this point, where the 45g of melted ice is now at a temperature of 76.07 degC,
all the 13 g of steam originally present has been converted to water at 100 degC.
The two different amounts of water at different temperatures will continue undergoing thermal
interaction until they finally come to the common temperature we're looking for.
By conservation of energy
heat absorbed by 45 g melted ice at 76.07 degC = heat lost by 13 g water at 100 degC
Now, let
Q3 = heat absorbed by 45g (0.045 kg) melted ice at 76.07 degC
Q4 = heat lost by 13 g (0.013 kg) water at 100 degC
Then
Q3 = m*(SHCW)*(dT) = m*(SHCW)*( TF - 72.07 ) = ( 0.045 ) ( 4190 ) ( TF - 72.07 )
which gives
(4) .......... Q3 = (188.55) ( TF - 72.07)
Similarly
Q4 = m*(SHCW)*(dT) = m*(SHCW)*( 100 - TF ) = ( 0.013) ( 4190 ) ( 100 - TF )
which yields
(5) .......... Q4 = (54.47 ) ( 100 - TF )
By conservation of energy, (4) and (5) are equal:
( 188.55 ) ( TF - 72.07) = ( 54.47 ) ( 100 - TF )
TF=78.33
The final temperature of the mixture will be 78.33degC with all the ice originally present melted.

Part (b)
Having a smaller amount of steam originally present, less heat energy will be available to melt
the 45g (0.045 kg) of ice. According to (2) of part (a)
(6) .......... Q2 = ( 0.0013 )( 2256 ) = 2.9328kJ
According to (1) of part (a), not all the ice will be melted but only a small amount.
In analogy with (1) of part (a) but with Q1 = 2.9328 kJ, we find that
( 2.9328 ) = m ( 333 )
which gives
m = ( 2.9328 ) / ( 333 ) = 0.0088 kg = 8.8 g
That is only 8.8 g of ice will be melted, with 45 g - 8.8g = 36. 2 g
of ice remaining at 0 degC.