The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.23 Ω . When the thermometer is inserted in a hot bath, the resistance of the platinum wire will be 5.795 Ω . Calculate the temperature of the bath.

Question

The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.23 Ω . When the thermometer is inserted in a hot bath, the resistance of the platinum wire will be 5.795 Ω . Calculate the temperature of the bath. (A)  3.456 °C  (B)  34.565 °C  (C)  345.65 °C  (D)  45.65 °C

Tardigrade Admin · Accepted Answer

R0=5 Ω , R100=5.23 Ω, Rt=5.795 Ω α=(R100-R0/R0×100)=(Rt-R0/R0× t) Therefore, t=(Rt-R0/R100-R0)×100 =((5.795-5/5.23-5))×100 =345.65 ° C

Q. The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5 Ω$ and at steam point is $5.23 Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire will be $5.795 Ω$ . Calculate the temperature of the bath.

$3.456^{\circ} C$

$34.565^{\circ} C$

$345.65^{\circ} C$

$45.65^{\circ} C$

$R_{0} = 5 Ω, R_{100} = 5.23 Ω, R_{t} = 5.795 Ω$
$α = \frac{R _{100} - R _{0}}{R _{0} \times 100} = \frac{R _{t} - R _{0}}{R _{0} \times t}$
$T h ere f ore, t = \frac{R _{t} - R _{0}}{R _{100} - R _{0}} \times 100$
$= (\frac{5.795 - 5}{5.23 - 5}) \times 100$
$= 345.65^{\circ} C$

Q. The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5Ω and at steam point is 5.23Ω . When the thermometer is inserted in a hot bath, the resistance of the platinum wire will be 5.795Ω . Calculate the temperature of the bath.

3.456∘C

34.565∘C

345.65∘C

45.65∘C