Day 12 Electric Potential 04/09/2014

Quiz:

100% Correct ~ LOL

Change Water Temperature by Electricity

In this lab, we heat up the water by coiled wired heater, which is about 42 cm. It attached to a 4.5 V battery, and place in the water for 10 minutes. In this lab, we were asked to find the temperature change with given wired heater length, time, and voltage.

First we calculate the resistance (R) of wired heater ( Power*Length / Area )then use the resistance to find the current (I ) (voltage / Resistance). Also Put in the uncertainty by differentiate two different power.

After solving current (I), we solver power by (P = I * V) With solved power we can solve delta T by Q = mc(delta T; however, power * t = Q. Use the value of uncertainty to plug in exact same formula that we were use will gives us the final uncertainty.

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With double voltage I = 2V/R => P = 2V/R * 2V = 4V^2/R (which is aquatically)
changing T = (p*t) / (m*c) = 4V^2t / (Rmc).
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This picture shows the temperature change from logger pro. The blue one is the 9 V and red is  4.5V

According to logger pro for the initial experiment the temperature started at about 23°C and concluded at about 25°C for a 2°C temperature change. The doubled voltage experiment started at 24.6°C and ended at 32.1°C for a 7.5°C temperature change.