Important Note : Please make sure you answer them correctly in good details. Als
ID: 1269740 • Letter: I
Question
Important Note : Please make sure you answer them correctly in good details. Also, please do NOT COPY AND PASTE PARAGRAHS FROM GOOGLE and try to solve it from your readings and experience.
Electric Joule Heating LAB :
Was the energy (Q) gained by the water greater than, equal to, or less than the energy lost by the resistor? Explain. How would one, over a number of trials, verify question #1 statistically? What is the purpose of the calorimeter in this experiment? Name some factors that could affect the accuracy of the measurements in this experiment. A resistor in an electrical circuit is analogous to a non-conservative mechanical force. What is this force? The current and voltage should remain fairly constant throughout this experiment. Why? The specific heat of the calorimeter is negligible but the specific heat of the ceramic resistor is about 800 J/kg- C Degree . Calculate Q total as the sum of Q water + Q resistor- How does this affect the outcome of this experiment?Explanation / Answer
1] the law energy states that the energy lost = energy gained
so the energy gained by the water = energy lost by the resistor
2]By checking the energy lost and gained at different time intervals
3]A calorimeter is an experimental device in which a chemical reaction or physical process takes place. The calorimeter is well-insulated so that, ideally, no heat enters or leaves the calorimeter from the surroundings. For this reason, any heat liberated by the reaction or process being studied must be picked up by the calorimeter and other substances in the calorimeter.
4]Heat loss through the calorimeter can effect the measurements
5]Electric force
6]so that the electric energy remains constant = VI
7] this one i am a bit confused
But you have to go by Q = mSdelta[T]
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