phys Now, connect the two Ieads of the power resistor to the variable power supp

Question

phys Now, connect the two Ieads of the power resistor to the variable power supply. Set up your system to provide 1.0 ampere of current (or the largest value obtainable with your resistor) by adjusting the current and voltage controls of the variable power supply. Once you have set the controls to give you the optimal voltage and current parameters. turn of the power supply, Ieaving the control settings as they were. Now. add about 50 ml of room temperature water to the cup and place the temperature sensor into the water. Set the Labquest to record temperature data for 10 minutes. Begin collecting data After 20 or 30 seconds turn on the power supply so that current is flowing through the resistor. At various times during the measurement shake the cup slightly to make sure the temperature measurement is close to the specific steady state temperature of the water/1 resistor system. After the 10-minutcs measurement. get a printed plot of temperature vs time so that you can readily determine and record the initial and final temperatures (to help with this you may want to print the table of data).

Explanation / Answer

EXPERIMENTALLY

time taken to raise temperature from 22.3 to 28.7 oC = 10mins = 600 secs

= 6.4 oC in 600 secs

so rate = 6.4 / 600 =0.10765

THEORITICALLY ,

R = 8.19 ohms

I = 1.03 A

heat lost = I2R = 8.688

Q= m*Sp*dT

m = mass of water , = 50 ml = 50 gram

specific heat of water = 4.186 J /g /oC

dT = 6.4 oC

so heat reqired = 50*4.186 * 6.4 = 1339.52 J

now , from resistor , heat required = I2Rt

=8.688* t

equating ,

8.688* t = 1339.52 J

=> t = 154.18 secs= 2.56 mins

dT/dt = 6.4 oC/154.18 =0.415 oC/sec

this variation is because experimentally there are heat losses to the environment ,hence the time required to reach the temp was more. theoritically , there is no loss of heat to enviroment , the heat is completely transferred to the water

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