Practice Problem #1 Solution.
Initial Conditions for objects are:
A: 4.0 kg Ag at 750 C with Specific Heat Capacity of
B: 4.0 kg Ice at 0 C with Specific Heat Capacity of
Are there any possible phase changes? If so, list the temperatures, substances and the
Latent Heats for each possible change.
Possible Phase change for the ice while still at 0 C
Heat Added
Temp of Ice
Heat Removed
0
Temp of Ag
750
Begin Phase Change
First
To melt Ice need Q = 4kg(Lmelt ice)
Q = 4kg (3.33x105 J/kg)
Q = 1,332,000 J
0
Remove energy from Ag to use to melt
ice.
Q = mcT
1,332,000 J = 4 kg (234 J/kgC)T
T = 1423.1  C
750-1423
This is lower than 0
degrees so
impossible.
Therefore: Some Ice melts but not all of it. If not all melts, the final temp must be 0 C. Now we
can solve the problem by calculating how much of the ice melts.
Heat Added
Temp of Ice
0
Begin Phase Change
To melt Ice using the heat from the Ag
Q = mice (3.33x105 J/kg)
702,000 = mice (3.33x105 J/kg)
m = 2.1 kg
0
Heat Removed
Temp of Ag
750
First: Remove energy from Ag to use to
melt ice.
Q = mcT
Q = 4 kg (234 J/kgC)  C
Q = 702,000 J
0
Practice Problem #2 Solution.
Initial Conditions for objects are:
A: 10.0 g Steam at 100 C with Specific Heat Capacity of
B: 50.0 g Ice at 0 C with Specific Heat Capacity of
Are there any possible phase changes? If so, list the temperatures, substances and the
Latent Heats for each possible change.
Possible Phase change for the steam while still at 100 C.
Possible Phase change for the ice while still at 0 C
I’ll just guess to try steam first since its L is higher for the phase change.
Heat Added
Temp of Ice
Heat Removed
0
Second:To melt Ice need Q = 4kg(L)
Q = 50g (79.7 cal/g)
Q = 3985 cal
Steam losses 5400 cal and ice uses 3985
of that to melt. Now we have water at 0
C. It will warm with the remaining Q.
Q to warm is:
0 + 28.3 =
5400 – 3985 cal
28.3  C
1415 cal = mcT
T = 28.3  C
Temp of Steam
100
Begin Phase Change
First: Remove energy from Steam to use
to melt ice.
Q = mL
5400 cal = 10 g (540 cal/g)
The steam did not
cool, only turned
into water!
100  C
Now there are two different amounts of water at different temps. Now we can solve the problem
by calculating final temp. I expect my final temp to be between 28.3 and 100.
Heat added = Heat removed
mcoolwatercT  mhotwatercT
cal
50 g (1 cal
gC )(T f  28.3C )  10 g (1 gC )(100  T f )
50T  1415  1000  10T
60T  2415
T f  40.2C
From Energy Equation:
cal
50 g (1 cal
gC )(T f  28.3C )  10 g (1 gC )(T f  100)  0
50T  1415  10T  1000  0
60T  2415
T f  40.2C
Method 1: I set this up so that both
sides are positive.
When I get to solving the equation,
the units were left out to make the
algebra clear. You should add units
back in when turning your work in.
Method 2: I set this up more like text.
Notice that the equations are really
the same!