Question

The resistance of a platinum wire of platinum resistance thermometer at the ice point is \[5\Omega \] and at the steam point is \[5.39\Omega \]. When a thermometer is inserted in a hot bath, the resistance of a platinum wire is \[5.795\Omega \]. Calculate the temperature of the bath.

Answer 1

Hint: We have given the value of resistance of platinum wire at two temperatures or points and the relation between resistance and temperature can be given by temperature coefficient of resistance. By using the formula of temperature coefficient of resistance we can find the temperature of a hot bath.
Formula used:
\[{{R}_{T}}={{R}_{0}}(1+\alpha T)\]

Complete answer:
The resistance of a material depends on a temperature which is given by the following formula
\[{{R}_{T}}={{R}_{0}}(1+\alpha T)\text{ }..................\text{(i)}\]
Where \[{{R}_{T}}\] is the resistance at any temperature T, \[{{R}_{0}}\] is the resistance at zero degree Celsius, α is the temperature coefficient of resistance and T is the temperature at which resistance has to be calculated.
Now we have given that platinum wire of platinum resistance temperature has resistance of \[5\Omega \] at ice point and \[5.39\Omega \] at steam point. The ice point is \[0{}^\circ C\] and steam point is \[100{}^\circ C\], by using above formula we can find temperature coefficient as we have given resistance at \[0{}^\circ C\] and at \[100{}^\circ C\]. Given data are
\[{{R}_{T}}=5.39\Omega, {{R}_{0}}=5\Omega \text{ and }T=100{}^\circ C\]
Substituting it in equation (i), we have
\[\begin{align}
  & 5.39=5(1+\alpha 100) \\
 & \Rightarrow (1+100\alpha )=\dfrac{5.39}{5} \\
 & \Rightarrow 100\alpha =1.078-1 \\
 & \Rightarrow \alpha =\dfrac{0.078}{100} \\
 & \Rightarrow \alpha =0.078\times {{10}^{-2}}=7.8\times {{10}^{-4}}{}^\circ {{C}^{-1}} \\
\end{align}\]
Now according to the question the platinum wire has resistance of \[5.795\Omega \]when the platinum resistance thermometer was inserted in a hot bath and we have to find the temperature of the bath. Let's say the temperature of the hot bath is T’ and as the wire is the same temperature coefficient will be the same. Therefore by using equation (i) we can find the temperature as resistance at T’ is given and we have calculated the temperature coefficient.
Substituting \[{{R}_{T'}}=5.795\Omega, {{R}_{0}}=5\Omega, \alpha =7.8\times {{10}^{-4}}{}^\circ {{C}^{-1}}\text{ and }T=T'\] in equation (i), we get
\[\begin{align}
  & 5.795=5\left( 1+\left( 7.8\times {{10}^{-4}} \right)T \right) \\
 & \Rightarrow 1+\left( 7.8\times {{10}^{-4}} \right)T=\dfrac{5.795}{5} \\
 & \Rightarrow \left( 7.8\times {{10}^{-4}} \right)T=1.195-1 \\
 & \Rightarrow T=\dfrac{0.195}{7.8\times {{10}^{-4}}} \\
 & \Rightarrow T=0.0203\times {{10}^{4}} \\
 & \Rightarrow T=203{}^\circ C \\
\end{align}\]
Hence the temperature of the hot bath is\[203{}^\circ C\].

Note:
Ice point refers to the point where pure water is at equilibrium and steam point is the point where pure water and vapour are at equilibrium. From the question itself we can see that resistance was changing with temperature therefore we have to use a relation between temperature and resistance to solve it. Conductors always have a positive temperature coefficient which means the resistance will increase with increase in temperature.