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 shouldve 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 58.78C

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

23C

46C

??

XXXXX

Water

.110 kg

75 C

46C

4187 J/kg/C

XXXX

ICE

.022 kg

0 C

46C

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 mustve 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 * CAl*(46-23)= 1748.986

                  CAl= 1748.986/(.016*23) = 4752.679348 Joules

 

 

How does this compare with the real specific heat?

 The real specific heat is 900 J/kgC

What is your percent error?

that gives me a percent error of (4752.6-900)/900 = 428 %!!

 

Why?

Well, I wasnt too far off! There was probably heat loss to the surroundings, and I didnt wait long enough. The cup only SHOULD have gained (.016 * 900 *23)=331.2 Joules of heatI 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.