The current is switched on at t=0 and allowed to run until the temperature is increased by 40 to 50℃. The temperature is recorded at regular intervals of about a minutes. If the maximum temperature reached after t is Tmax, then the energy supplied by the battery to the liquid is
heat supplies=Vit
where I is the current in the heater and V the voltage across it. The heat absorbed by the liquid and the calorimeter is
heat absorbed=mc(Tmax–T)+C(Tmax-T)
Equating the two quantities of heat allows us to determine c.
EQUIPMENTS:
Calorimeter, digital balance, ice cubes, thermometer, warm water, and paper towel.
METHOD:
- Prepare the equipment.
- Measure the mass of container, and put little water into the container.
- Put the boiling water into the container when it reach the half volume of container.
- measure the initial temperature of water, and record data.
- Measure the container with water, then calculate the mass of water.
- Put the ice into the water.
- After the disappear of the ice, measure the mass of the caontainer and calculate the mass of ice.
- Measure the final temperature of water and record data.
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The relevant data was recorded and the value of Li was calculate.
DATA COLLECTION:
DATA ANALYSIS:
The specific heat capacity of water: 4.18E3 Jkg-1K-1
The speaific heat capacity of ice: 2.1E3 Jkg-1K-1
CwMw(Tf-Ti)+CAlMAl(Tf-Ti)=MiLi+CwMiTf
4.18×87.8*(58-46)+0.91*29.3*(58-46)= 8.9*Li+4.18*8.9*46
Li≈338.5J/g
CONCLUSION:
Our conclusion of the latent heat of fusion of the ice is about 338.5J/g.
EVALUATION:
I think there are some heat transfer to air. And when we measure the initial temperature, the heat transferred during the time we waiting.
CONFERENCE:
<Physics for the IB diploma>
Fourth Edition
K.A. Tsokos
