The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco
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Answer: The students will determine the two fixed points of the thermometer:
Lower fixed point = 0 degree Celsius
Upper fixed point = 100 degree Celsius
Then divide the thermometer with equal intervals
The room temperature will be the point at which the themometric substance remains constant when rising from ice point.
Explanation:
Apparatus available:
Unmarked thermometer
250 cm3 glass beaker
crushed ice
water
heatproof mat
clamp, boss and stand
meter rule
Added apparatus
Bunsen burner
Stirrer
Method
The students will determine the two fixed points of the thermometer:
Lower fixed point = 0 degree Celsius
Upper fixed point = 100 degree Celsius
Then divide the thermometer with equal intervals
Procedures
Set up the apparatus of illustrated in the attached figure.
Immerse the unmarked thermometer into the ice in the beaker.
When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.
You may add little water and continue to stir gently.
The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value
Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point
This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.
Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.
