Heat and temperature | NEB Class 11 Physics

1. Temperature

Temperature is the average kinetic energy at all molecules of present in that body. It is measured by thermometer and its SI unit is Kelvin.

2. Heat

Heat is defined as the sum of all molecules of a substance kinetic energy of all molecules of that substance.

3. Effect of heat

When heat is applied to a body, it brings the following changes;

1. Change in temperature
2. Change in volume
3. Change of state of matter
4. Electric effect
5. Generation of light energy
6. Change in resistance

4. Thermal Equilibrium

Two bodies are said to be in thermal equilibrium with each other when no heat flows from one body to another when they are brought in thermal contact.

5. Zeroth law of Thermodynamics

It states that if two system are separately in thermal equilibrium with a third system then they must be in thermal equilibrium with each other.

Thermodynamic System: The system which can be explained by thermodynamic parameters (Temperature, Pressure, Volume, heat exchange).

6. Thermometry

Thermometry is the science of temperature and its measurement.

Three main steps of thermometry:

1. Construction
2. Calibration
3. Sensitivity

Temperature scale and their relation

Relation between them:

$$ \frac{C-0}{100}=\frac{F-32}{180}=\frac{K-273}{100}=\frac{R-0}{80} $$

Numerical problems

At what temperature will the Kelvin scale reading double the fahrenheit reading.

Solution:

Let the temperature in Kelvin be \((k)\)

By question: \(K = 2F\)

We know,

$$ \frac{F - 32}{180} = \frac{K - 273}{100} $$

or, $$ \frac{F - 32}{180} = \frac{2F - 273}{100} $$

or, \(200f - 3200 = 360f - 49140\)

or, \(45940 = 260f\)

\(f = 176.69\)

Also, \(2f = 176.69 \times 2 = 353.38\)

Hence, at \(176.69^{\circ}F\) or \(353.38K\) temperature reading will be double

1. Convert \(30^{\circ}C\) into kelvin scale fahrenheit. Scale.

Solution:

Given, Temperature in celsius (C) = \(30^{\circ}C\)

Temperature in fahrenheit (F) = ?

We know,

$$ \frac{C-0}{100} = \frac{f-32}{180} $$

or, $$ \frac{30}{100} = \frac{f-32}{180} $$

or, $$ \frac{30 \times 180}{100} + 32 = F $$

\(F = 86^{\circ}F\)

Hence, \(30^{\circ}C = 86^{\circ}F\)

2. At what temperature will the celsius reading double of fahrenheit reading.

Solution:

Let the temperature in celsius be \((2x)\) and fahrenheit be \((x)\)

we know,

$$ \frac{C-0}{100} = \frac{F-32}{180} $$

or,

$$ \frac{2x}{100} = \frac{x-32}{180} $$

or,

\(360x = 100x - 3200\)

$$ \therefore x = -12.307^{\circ}F $$

Hence, at \(-12.307^{\circ}F\), celsius reading will be double the fahrenheit reading.

3. At what temperature Celsius scale reading coincides with Fahrenheit scale.

Solution:

Let the temperature in Celsius and Fahrenheit scale be \(x\)

We know,

$$ \frac{c-0}{100}=\frac{f-32}{180} $$

or, \(180x = 100x - 3200\)

$$ \therefore x = -40^{\circ}C $$ or \(-40^{\circ}F\)

At \(-40^{\circ}C\) celsius Scale coincides with Fahrenheit Scale.

4. A faulty thermometer has its fixed point marked at 2 and 98. What is the correct temperature on the Celsius scale when the thermometer reads \(20^{\circ}C\).

Solution:

Given, L.F.P of faulty thermometer \((LFP_{f}) = 2^{\circ}C\)

U.F.P of faulty thermometer \((UFP_{f}) = 98^{\circ}C\)

Temperature measured in faulty thermometer \((C_{f}) = 20^{\circ}C\)

LFP of Correct thermometer (LFP) = \(0^{\circ}C\)

UFP of Correct thermometer (UFP) = \(100^{\circ}C\)

Correct reading in celsius (C) = ?

We know,

$$ \frac{C_{f}-LFP_{f}}{UFP_{f}-LFP_{f}} = \frac{C-LFP}{UFP-LFP} $$

or, $$ \frac{20-2}{98-2} = \frac{C-0}{100-0} $$

or $$ \frac{18}{96} = \frac{c}{100} $$

$$ \therefore C=18.75^{\circ}C $$ Hence, the correct reading in celsius is \(18.75^{\circ}C\).

5. At what point of thermometric scale does kelvin scale reading coincide with fahrenheit scale reading.

Solution:

Let the temperature in kelvin scale and fahrenheit scale be \(x\):

We know,

$$ \frac{f - 32}{180} = \frac{k - 273}{100} $$

or, $$ \frac{x - 32}{180} = \frac{x - 273}{100} $$

or, \(100x - 3200 = 180x - 49140\)

\(x = 574.25\) K or \(574.25^{\circ}F\)

6. Fahrenheit thermometer reads \(99^{\circ}\) when a standard centigrade thermometer reads \(37^{\circ}C\). Find the error in fahrenheit scale.

Solution:

Fahrenheit reading = \(99^{\circ}F\)

Standard Centigrade reading = \(37^{\circ}C\)

Equivalent temperature of \(37^{\circ}C\) into fahrenheit is

$$ \frac{c-0}{100}=\frac{f-32}{180} $$

or, $$ \frac{37}{100} = \frac{f-32}{180} $$

or, \(6660 = 100F - 3200\)

$$ \therefore f = 98.6^{\circ} $$

Now,

fault in Fahrenheit Scale = \(99^{\circ}F - 98.6^{\circ}F\)

\(= 0.4^{\circ}F\)

7. You work in a materials testing lab and your boss tells you to increase the temperature of a sample by \(40^{\circ}C\). The only thermometer you can find at your work bench reads in \( ^{\circ}F\). If the initial temperature of the sample is \(68.2^{\circ}F\) what is the final temperature in \( ^{\circ}F\) when the desired temp. is increased.

Solution:

Initial temperature in \(f = 68.2^{\circ}F\)

Required change in temperature = \(40^{\circ}C\)

Final temp in \(f = ?\)

we know,

100 parts in \( ^{\circ}C\) = 180 parts in \( ^{\circ}F\)

or 1 parts in \( ^{\circ}C\) = \(\frac{180}{100}\) parts in \( ^{\circ}F\)

or, 40 parts in \( ^{\circ}C\) = \(\frac{180}{100} \times 40\) parts in \( ^{\circ}F\)

\(=72^{\circ}F\)

Now, Final temperature = \(68.2^{\circ}F + 72^{\circ}F\)

\(=140.2^{\circ}F\)

8. What are the temperature in Celsius and Fahrenheit scale when \(\frac{2}{3}\) of fahrenheit scale is \(\frac{1}{2}\) of celsius scale?

Solution:

By Question:

$$ \frac{2}{3}f = \frac{1}{2}c $$

or, $$ f = \frac{3}{4}c $$

$$ \frac{c-0}{100}=\frac{f-32}{180} $$

or, $$ \frac{c}{100} = \frac{0.75c - 32}{180} $$

$$ \therefore c = -30.48 $$

or $$ F = \frac{3}{4}\times(-30.48) $$

$$ f = -22.86^{\circ}F $$

9. A celsius scale reads \(30^{\circ}C\), when a standard kelvin scale reads 300K. What is the error in celsius scale?

Solution:

Here, Reading in \(c = 30^{\circ}C\)

Standard Kelvin Scale reading = 300K

Corresponding temperature of 300k into \( ^{\circ}C\).

we know $$ \frac{c - 0}{100} = \frac{k - 273}{100} $$

or, $$ \frac{c}{100} = \frac{300-273}{100} $$

or, \(c = 27^{\circ}C\)

Now, Error in celsius scale reading = \(30^{\circ}C - 27^{\circ}C\)

\(= 3^{\circ}C\)

10. A thermometer has wrong calibration. It reads the melting point of ice \(-10^{\circ}C\). It reads \(60^{\circ}C\) in place of \(50^{\circ}C\). What is the temperature of boiling point of water, on this scale?

Here, Lower fixed point of faulty thermometer \((LFP_{f}) = -10^{\circ}C\)

Reading of faulty thermometer \((Cf)\) = \(60^{\circ}C\)

Correct reading of thermometer \((C)\) = \(50^{\circ}C\)

Upper fixed point of faulty thermometer \((UFP_{f}) = ?\)

we know,

$$ \frac{C_{f} - LFP_{f}}{UFP_{f} - LFP_{f}} = \frac{C - 0}{UFP - LFP} $$

or, $$ \frac{60-(-10)}{x-(-10)}=\frac{50-0}{100-0} $$

or, $$ \frac{70}{x+10} = \frac{50}{100} $$

or \(7000 = 50(x + 10)\)

or $$ \frac{7000 - 500}{50} = x $$

$$ \therefore x = 130^{\circ}C $$

11. In an arbitrary scale of temperature, water freezes at \(40^{\circ}C\) and boils at \(290^{\circ}C\). Find the boiling point of a liquid in this scale if it boils at \(62^{\circ}C\).

Solution:

LFP of faulty thermometer (LFPf) = \(40^{\circ}C\)

UFP of faulty thermometer (UFPf) = \(290^{\circ}C\)

Boiling point of liquid in faulty thermometer (Cf) = ?

Accurate boiling point of liquid in thermometer (C) = \(62^{\circ}C\)

We know,

$$ \frac{C-0}{UFP-LFP} = \frac{CF-LFP_f}{UFP_f-LFP_f} $$

or, $$ \frac{62}{100} = \frac{Cf-40}{290-40} $$

or, $$ \frac{62}{100} = \frac{Cf-40}{250} $$

or, \(15500 = 100 Cf - 4000\)

$$ \therefore $$ Cf = \(195^{\circ}C\)

12. The distance between LFP and UFP is 80 cm. Find the temperature on the celsius scale if the mercury level rises to height of 10.4 cm, above the lower fixed point.

Solution:

Distance between lower fixed LFP and UFP = 80 cm

we know,

80 cm = 100 parts in celsius thermometer

1 cm = \(\frac{100}{80}\) parts

10.4 cm = \(\frac{100}{80} \times 10.4\) parts

= 13 parts.

Here, 1 part = \(1^{\circ}C\). So 13 parts = \(13^{\circ}C\).

Original Method:

Height of 80 cm in thermometer = \(100^{\circ}C\)

Height of 1 cm in thermometer = \(\frac{100}{80}^{\circ}C\)

Height of 10.4 cm in thermometer = \(\frac{100}{80} \times 10.4^{\circ}C\)

= \(13^{\circ}C\)

Short Questions

1. Does temperature depend on the amount of heat?

   → Yes, temperature depends on the amount of heat supplied, nature and mass of substance.

2. Can an object be hotter than another if they are at the same temperature? Explain

   No, one object cannot be hotter than another if they are at the same temperature.

3. If you have a bucket of cold water and a cup of hot tea which one of them have greater temperature? What about heat.