Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Temperature. Click on this to access more such competitive quizzes on Thermal Physics.

Temperature Quiz 2

physics

 A glass thermometer has some error in its markings. If the $LFP$ is $-{{2}^{\circ }}C$ and the $UFP$ is ${{118}^{\circ }}C$ . Then calculate the error occurring when the actual temperature is ${{28}^{\circ }}C$. 

 Calculate the corresponding faulty temperature and subtract the accurate value from it. 

  $\text{We all know that for temperature scales we can write;}\\\dfrac{x-LFP}{UFP-LFP}=K\\\text{So assuming the faulty scale as a random scale and comparing it with the accurate Celsius scale we can write;}\\\dfrac{28}{100}=\dfrac{X-\left( -2 \right)}{118-\left( -2 \right)} \\\Rightarrow \dfrac{28}{100}=\dfrac{X+2}{120} \\\Rightarrow 28\times 12=10X+20 \\\Rightarrow 336-20=10X \\\Rightarrow X=\dfrac{316}{10} \\\Rightarrow X={{31.6}^{\circ }}C \\\text{Now the error is the difference in accurate and measured values;}\\\Delta \theta =X-C \\\Rightarrow \Delta \theta =31.6-28 \\\therefore \Delta \theta ={{3.6}^{\circ }}C$

 For a bag full of bricks if the bag is at a temperature of ${{128}^{\circ }}C$ , what is the temperature of the brick in the topmost layer? 

 Recall the zeroth law of thermodynamics that defines temperature. 

  According to the zeroth law of thermodynamics that defines temperature, if one body is in thermal equilibrium with another body and this body is in thermal equilibrium with some third body then the third body is at thermal equilibrium with the first body. Same goes for a long chain of bodies. The heat will transfer until all the bodies will reach the same temperature. Thus the brick at top will be at the same temperature ${{128}^{\circ }}C$.

 For a random temperature scale the ice point and steam point is ${{12}^{\circ }}X$ and ${{132}^{\circ }}X$ respectively. What is the relation between this scale and the Fahrenheit scale? 

 The ratio of temperature difference in case of Fahrenheit scale will be equal to the ratio of temperature difference for the random scale. 

 $\text{The ratio of change in temperature for all the scales is constant.}\\\text{We know that the comparison of the two scales can be done as;}\\\dfrac{F-32}{180}=\dfrac{X-12}{132-12} \\\Rightarrow \dfrac{F-32}{180}=\dfrac{X-12}{120} \\\Rightarrow F-32=\dfrac{3}{2}X-18 \\\therefore F=\dfrac{3}{2}X+14$

 An ideal gas thermometer has a random scale. If pressures at $LFP$ and $UFP$ are ${{P}_{0}}$ and $3{{P}_{0}}$ respectively. What should be the relation of temperature in degrees Celsius with the pressure of the gas? 

 Assume an arbitrary temperature variable in the random scale and relate it to a corresponding temperature variable in degree Celsius. 

 $\text{Let P be the pressure of the ideal gas at any arbitrary temperature}\,{{T}^{\circ }}X.\\\text{Since this gas is an ideal gas we can write;}\\\dfrac{T}{P}=\dfrac{LFP}{{{P}_{L}}}=\dfrac{UFP}{{{P}_{U}}}\\\text{So we get;}\\LFP=\dfrac{T{{P}_{L}}}{P}$ and $UFP=\dfrac{T{{P}_{U}}}{P}\\\text{Let}\,{{C}^{\circ }}C\,\text{be the corresponding temperature of}\,{{T}^{\circ }}X.\\\text{Now if we compare the temperature readings by the random scale and the Celsius scale we get;}\\\dfrac{C}{100}=\dfrac{T-LFP}{UFP-LFP}\\\text{Substituting the values we get;}\\\dfrac{C}{100}=\dfrac{T-\dfrac{T{{P}_{L}}}{P}}{\dfrac{T{{P}_{U}}-T{{P}_{L}}}{P}} \\\Rightarrow \dfrac{C}{100}=\dfrac{P-{{P}_{L}}}{{{P}_{U}}-{{P}_{L}}} \\\Rightarrow C=100\left( \dfrac{P-{{P}_{L}}}{{{P}_{U}}-{{P}_{L}}} \right)\\\therefore C=100\left( \dfrac{P-{{P}_{0}}}{ 2{{P}_{0}}} \right)$

 An ideal gas thermometer works on a gas $G$ . The scale is random and the ice point is ${{T}_{1}}$ and the pressure at the ice point is ${{P}_{1}}$ . What must be the steam point if pressure of gas at ${{25}^{\circ }}C$ is ${{P}_{2}}$ 

 Find the temperature and compare it with the given Celsius reading. 

 $\text{Now the gas is ideal so the pressure temperature ratio is constant. Thus we can say;}\\\dfrac{{{T}_{1}}}{{{P}_{1}}}=\dfrac{{{T}_{2}}}{{{P}_{2}}} \\\Rightarrow {{T}_{2}}=\dfrac{{{P}_{2}}{{T}_{1}}}{{{P}_{1}}} \\\text{Now comparing the Celsius scale with the random scale we get;}\\\dfrac{C}{100}=\dfrac{{{T}_{2}}-{{T}_{1}}}{UFP-{{T}_{1}}} \\\Rightarrow\dfrac{25}{100}=\dfrac{{{T}_{2}}-{{T}_{1}}}{UFP-{{T}_{1}}} \\\Rightarrow UFP-{{T}_{1}}=4\left( {{T}_{2}}-{{T}_{1}} \right) \\\Rightarrow UFP=4{{T}_{2}}-3{{T}_{1}} \\\Rightarrow UFP=\dfrac{4{{P}_{2}}{{T}_{1}}}{{{P}_{1}}}-3{{T}_{1}} \\\therefore UFP=\left( \dfrac{4{{P}_{2}}}{{{P}_{1}}}-3 \right){{T}_{1}}^{\circ }X$