Data collection and calculation :
Zinc :
From the Conservation of energy, we can know:
Heat lost by Copper = Heat gained by cold water
So the Zinc : cm(T2-T3) = Cold water: cm(T3-T1)
Trial 1:
c(water)=4200℃
M(water)=M3-M2=0.040-0.002=0.038kg (±0.001kg)
C(zinc)×0.054×(86.0-26.0)=4200×0.038×(26.0-20.0)
C(zinc)=296℃
Trial 2:
c(water)=4200℃
m(water)=0.041-0.002=0.039 kg (±0.001kg)
C(zinc)×0.054×(89-25)=4200×0.039×(25-19)
C(zinc)=283℃
Trial 3:
M(water)=m3-m2=0.046-0.002=0.044 kg (±0.001kg)
m(water)=0.046-0.002=0.044 kg (±0.001kg)
C(zinc)×0.054×(79-23)=4200×0.039×(23-18)
C(zinc)=271 ℃
The average specific heat of
zinc =(270.833+283.375+295.556)/3=283.255℃
So the percentage of error is：|(283.255-380)/380|×100%=25.46%
Aluminum :
the Aluminum : cm(T3-T2) = Cold water: cm(T3-T1)
Trial 1:
c(water)=4200℃
M(water)=M3-M2=0.046-0.002=0.044kg (±0.001kg)
C(Al)×0.020×(84.0-23.0)=4200×(0.046-0.002)×(23.0-18.0)
C(Al)=924 ℃
Trial 2:
c(water)=4200℃
m(water)= 0.042-0.002=0.040kg (±0.001kg)
C(Al)×0.020×(87.0-30.0)=4200×0.040×(30.0-21.0)
C(Al)=1326℃
Trial 3:
M(water)=m3-m2=0.041 kg (±0.001kg)
C(Al)×0.020×(74.0-24.0)=4200×0.041×(24.0-17.0)
C(Al)=1205℃
The average specific heat of
Aluminum =(924+1326+1205)/3=1152℃
So the percentage of error is：
|(1152-910)/910|×100%=26.59%
Copper :
the Copper : cm(T3-T2) = Cold water: cm(T3-T1)
Trial 1:
c(water)=4200℃
M(water)=M3-M2=0.039kg (±0.001kg)
C(copper)×0.062×(89.0-25.0 )=4200×0.039×6
C(Al)= 248℃
Trial 2:
c(water)=4200℃
m(water)=0.038 kg (±0.001kg)
C(Copper)×0.062×(89.0-25.0)=4200×0.038×(25.0-18.0)
C(Copper)=282℃
Trial 3:
M(water)=m3-m2= 0.037kg (±0.001kg)
C(Copper)×0.062×(85.0-18.0)=4200×0.037×(25.0-18.0)
C(Copper)=262℃
The average specific heat of
Copper =(248+282+262)/3=264℃
So the percentage of error is：
|(264-385)/385|×100%=31.43%
Evaluation:
Limitation:
In the experiment of the three numbers of specific heat are both higher than real specific heat of those three. Why like this? I think is :
- We can’t move the metal from boil water to cold water immediately. When the metal on the air, it will lose heat because of the temperature of metal is much higher than air. So the temperature of the metal will decrease.
- The thermometer is cold, when put into the hot water, the heat of the hot water will transfer to the thermometer, so the temperature will decrease.
- Sometimes not parallel to see the temperature, it will cause the value incorrect.
- when we use thermometer sometimes it touched the bottom of the beaker. So the temperature of metal will be hotter.
Error:
We got the Parallax error during measure the temperature of using thermometer. Because if we not parallel with the top of the mercury in the thermometer, it will makes the result to be a little bit different. But not affect a lot.
Improvement:
1. When you put the hot metal into the container of the cold water, you must be careful not to let the metal touch the container and make the container be occluded immediately so that the heat can be transfer into the cold water.
2 The big problem is the heat losing on the way that the metal is transferred into the container of the cold water. But I think if we can do this experiment in vacuum the heat lost will be decrease and the data we collected will more and more close to the data in theory.
