Answer to Question #162669 in General Chemistry for Mazwi Casbeth

Question

A copper-constant thermocouple generates voltage of 4.75*10^-3 boys when the temperature of the junction is 110.0°c and the reference junction is kept at a temperature of 0.0°c. if the voltage is proportional to the difference in temperature between the junction. What is the temperature of the hot junction when the voltage is 3.32*10^-3 volts

Answer 1

Q162669

A copper-constant thermocouple generates a voltage of 4.75*10^-3 volts when the temperature of the junction is 110.0°c and the reference junction is kept at a temperature of 0.0°c. if the voltage is proportional to the difference in temperature between the junction. What is the temperature of the hot junction when the voltage is 3.32*10^-3 volts?

Solution:

In the question, we are told that the voltage is proportional to the difference in temperature between the junction.

Voltage (V) ∝ ∆T

Voltage(V) = k ∆T,

where 'k' is the proportionality constant.

∆T is the difference in temperature between the junction.

Step 1: To find the proportionality constant, k.

A voltage of 4.75*10^-3 volts is generated when the temperature of the junction is 110.0°C and the reference junction is 0.0°C.

plug voltage(V) = 4.75*10^-3 volts and

∆T =  110.0°C - 0.0°C =  110.0°C in the formula we have

4.75*10^-3 volts = k *  110.0°C

divide both the side by  110.0°C, we have

$\frac{4.75*10^{-3} volts}{ 110.0°C} = \frac{k \space * \space 110.0°C}{110.0°C}$

$\frac{4.75*10^{-3} volts}{ 110.0°C} = \space k$

k = 4.32 * 10^-5 volts/ °C

Step 2: To find the temperature of the hot junction when the voltage is 3.32*10^-3 volts.

plug voltage = 3.32 * 10^-3 volts and k = 4.32 * 10^-5 volts/ °C in the formula

and find the temperature difference required for obtaining this voltage.

Voltage(V) = k ∆T,

3.32 * 10^-3 volts = 4.32 * 10^-5 volts/ °C * ∆T

divide both the side by 4.32 * 10^-5 volts/ °C we have

$\frac{3.32 * 10^{-3} volts}{ 4.32 * 10^{-5} volts/ °C} = \frac{4.32 * 10^{-5} volts/ °C * ∆T}{4.32 * 10^{-5} volts/ °C }$

76.88⁰C = ∆T;

The temperature of the reference junction is kept the same that is 0.0 °C.

The temperature of the hot junction - the temperature of the reference junction

= ∆T = 76.88⁰C

The temperature of the hot junction - 0.0 °C = 76.88⁰C

The temperature of the hot junction = 76.88⁰C.

In question, we are given the temperature in 1 decimal place, so our final answer must also be written in 1 decimal place.

Hence we must keep the temperature of the hot junction at 76.9 ⁰C for obtaining

3.32*10^-3 volts.