NAME________ SAMPLE!!
HEAT LAB 05:
PART ONE
PREDICT THE
TEMPERATURE!!
Take an amount of
Hot Water and half the amount of Cold Water.
Predict what you
think the final temperature will be!
Mix Them!
OBJECT |
MASS |
StartingTemp |
Final Temp |
SpecificHeat |
Hot Water |
.213 kg |
78 °C |
55 |
4187 J/kg/°C |
Cold Water |
.098 kg |
17 °C |
55 |
4187 J/kg/°C |
How much heat did
the Hot Water lose?
ΔQ
= mCΔT =
.213kg*
(4187)*(78-55)=
.213kg
*4187*(23°)=20, 512 Joules
How much heat did
the Cold Water gain?
ΔQ=mCΔT=
.098*4187*(55-17)=
.098*4187*(38°)=
15, 592 Joules
Are these the
same? Should they be? What should’ve the final temperature been? What is your
percent error? Why? Explain what happened in terms of the molecules and heat
transfer.
We lost 20,512-15,592 Joules of Heat=
4920 Joules of heat lost to air
The final
temperature for this should have been 58.78°C
mCΔT =
mCΔT
.213 (4187)(78
– Tf) = .098 (4187) (Tf-17)
891.831
(78-Tf) = 410.326 (Tf-17)
69562.8
– 891.8 Tf = 410.3Tf -6975.5
76538.3 =
1302.1Tf
Tf =
76538.3/1302.1 = 58.78 °C
Percent
Error = |Theor-Actual| / Theor =
(58.78-55)/58.75
= 3.78/58.78= 6.43 % error
PART TWO:
Determine the
specific heat of aluminum!
Place an amount of
HOT water in an aluminum calorimeter, along with an amount of an ice cube. When
the ice melts, determine the final temperature and the specific heat of the
calorimeter.
OBJECT |
MASS |
START TEMP |
FINAL TEMP |
SPECIFIC HEAT |
Heat of Fusion |
Al Cup |
.016 kg |
23°C |
46°C |
?? |
XXXXX |
Water |
.110 kg |
75 ° C |
46°C |
4187 J/kg/°C |
XXXX |
ICE |
.022 kg |
0 °C |
46°C |
2090 J/kg/°C |
335,000 J/kg |
How much energy
did the hot water lose?
The hot
water lost ΔQ= mCΔT = .110 * 4187*(75-46)
=.110*4187*29=
=13356
Joules
How much energy
did the ice gain by melting?
By melting
(solid to liquid) = m Hf = .022 * 335,000=
=7370
Joules
How much energy
did the ice gain by heating up (as water)?
As Ice Water
it gained ΔQ= mCΔT = .022 * 4187*(46-0)
=.022*4187*46=
=4237.244
Joules
How much energy
must’ve the Al cup gained?
Hot Water
Lost = Ice Gained (Melting + Heating) + Al Cup gained
13356 =
(7370 + 4237.244) + Al Cup gained
Al Cup
gained = 13356.23 – (7370+4237.244) = 1748.986 Joules
What is the
specific heat of Al according to your lab?
It gained
ΔQ= mCΔT = .016 * C_{Al}*(46-23)= 1748.986
C_{Al}=
1748.986/(.016*23) = 4752.679348 Joules
How does this
compare with the real specific heat?
The real specific heat is 900 J/kg°C
What is your
percent error?
that gives
me a percent error of (4752.6-900)/900 = 428 %!!
Why?
Well, I
wasn’t too far off! There was probably heat loss to the surroundings, and I
didn’t wait long enough…. The cup only SHOULD have gained (.016 * 900
*23)=331.2 Joules of heat…I calculated it gained 1748 Joules, so 1415 Joules
was lost to the surroundings.
The real
final temperature should have been around 48.5 °C
.110 *
4187*(75-Tf) = .022 * 335,000 + .022 * 4187*(Tf-0) + .016 * 900*(Tf-23)
algebra or
math solver shows the real final temp….
Explain what
happened in terms of the molecules and heat transfer.