Measuring Thermal Conductivity

The thermal efficiency apparatus contains a "Peltier Device". This device generates a voltage when it experiences a temperature difference at its ends. We can measure the rate at which it conducts heat from a hot reservoir to a cold reservoir by measuring the temperature difference and the power consumed from the hot reservoir.

We will need the Thermal Efficiency Apparatus, a 12V 2.5A DC power supply, 5 kg ice, a bucket, water, an ohmmeter, an ammeter, a voltmeter and wire.

Name:

Lab Partners:

When entering numeric data, use exponentials: ie., 1.6 * 10^-19 = 1.6E-19.

Procedure

Place the ice in the bucket and cover with water. Connect the pump power supply and place the tubes in the water-ice bucket.

Connect the meters and the power supply as shown (open mode:)

Turn the power supply on, and set it for 5 volts. Flip the toggle switch near the ohmmeter connection to the right to record Th (to the left for Tc.) Th should not be above 80 degrees C for more than 5 minutes. Read the temperature from the chart on the apparatus, which translates the thermistor resistance into temperature. Interpolate where necessary:
T = T_low + ( T_high - T_low ) * ( R_low - R ) / ( R_low - R_high )
For example, suppose the thermistor reads 89.5 kW. This will give a temperature between 27 and 28 C, because the chart gives a resistance of 91.1 kW for 27 C ( T_low ) and 87.0 kW for 28 C ( T_high ). To compute the temperature corresponding to 89.5 kW, we have
T = 27 + ( 91.1 - 89.5 ) / ( 91.1 - 87.0 )
= 27.39 C.

Measure Th and Tc repeatedly, until they come to equilibrium. This may take 5 to 10 minutes. You do not need to record the interim values. Record the equilibrium temperatures and the voltage and current:

R_{c, 5} = kW R_{h, 5} = kW
R_{Low, c, 5} = kW R_{Low, h, 5} = kW
R_{High, c, 5} = kW R_{High, h, 5} = kW
T_{Low, c, 5} = C T_{Low, h, 5} = C
T_{High, c, 5} = C T_{High, h, 5} = C
T_{c, 5} = C T_{h, 5} = C

V₅ = V I₅ = A

Repeat steps 3 and 4 for power supply voltages of 4 and 3 volts:

R_{c, 4} = kW R_{h, 4} = kW
R_{Low, c, 4} = kW R_{Low, h, 4} = kW
R_{High, c, 4} = kW R_{High, h, 4} = kW
T_{Low, c, 4} = C T_{Low, h, 4} = C
T_{High, c, 4} = C T_{High, h, 4} = C
T_{c, 4} = C T_{h, 4} = C

V₄ = V I₄ = A

R_{c, 3} = kW R_{h, 3} = kW
R_{Low, c, 3} = kW R_{Low, h, 3} = kW
R_{High, c, 3} = kW R_{High, h, 3} = kW
T_{Low, c, 3} = C T_{Low, h, 3} = C
T_{High, c, 3} = C T_{High, h, 3} = C
T_{c, 3} = C T_{h, 3} = C

V₃ = V I₃ = A

Analysis

There are 142 elements in the Peltier device which conduct heat in parallel between the hot and cold reservoirs. Each element has a ratio of distance to cross sectional area of 8.46 / cm. Calculate the thermal conductivity for each voltage from the data in steps 4 and 5:
k_i = V_i * I_i * ( 8.46 / 142 ) / ( T_{h, i} - T_{c, i} )
k₅ = J / s cm C
k₄ = J / s cm C
k₃ = J / s cm C

Which value should be the most accurate; why?

Look up the value for the thermal conductivites of copper and various glasses and ceramics:

Compare with the results of your experiment. What do you think the Peltier device is made of?

Please send comments or suggestions to the author.

R_{c, 5} = kW				R_{h, 5} = kW
R_{Low, c, 5} = kW				R_{Low, h, 5} = kW
R_{High, c, 5} = kW				R_{High, h, 5} = kW
T_{Low, c, 5} = C				T_{Low, h, 5} = C
T_{High, c, 5} = C				T_{High, h, 5} = C
T_{c, 5} = C				T_{h, 5} = C

R_{c, 4} = kW				R_{h, 4} = kW
R_{Low, c, 4} = kW				R_{Low, h, 4} = kW
R_{High, c, 4} = kW				R_{High, h, 4} = kW
T_{Low, c, 4} = C				T_{Low, h, 4} = C
T_{High, c, 4} = C				T_{High, h, 4} = C
T_{c, 4} = C				T_{h, 4} = C

R_{c, 3} = kW				R_{h, 3} = kW
R_{Low, c, 3} = kW				R_{Low, h, 3} = kW
R_{High, c, 3} = kW				R_{High, h, 3} = kW
T_{Low, c, 3} = C				T_{Low, h, 3} = C
T_{High, c, 3} = C				T_{High, h, 3} = C
T_{c, 3} = C				T_{h, 3} = C