The energy that is being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat.
Let us now find out the equation for the three measurements.
Clinical Thermometer
Q1: The quantity of heat energy required to change the temperature of one gram of water by one degree Celsius is known as
(a) 1 Joule
(b) One Kilojoule
(c) 1 Calorie
(d) 1 Ampere
Ans: (c)
Explanation: The amount of heat required to raise the temperature of one gram of water through 1° (from 14.5°C to 15.5°C), is called one calorie.
The measurement of temperature, a fundamental physical property, is achieved through specific scales using instruments called thermometers. These thermometers are calibrated to assign a numerical value to a given temperature.
Temperature measurement is typically expressed in units like Kelvin (K), degree Celsius (°C), and degree Fahrenheit (°F), with Kelvin (K) recognized as the SI unit. Various physical properties of materials that change with temperature, such as liquid volume, gas pressure, or metal resistance, form the basis of thermometer construction.
To define any temperature scale, two reference points are essential. For water, these are the ice point (where water freezes) and the steam point (where water boils) under standard pressure. On the Fahrenheit scale, water freezes at 32°F and boils at 212°F, whereas on the Celsius scale, water freezes at 0°C and boils at 100°C. The Celsius scale has 100 equal intervals between these reference points, while the Fahrenheit scale has 180.
Celsius versus Fahrenheit graph
The relationship between Fahrenheit (TF) and Celsius (TC) temperatures can be expressed as:
For Kelvin and Celsius, as they share the same unit size, the relation is:
Different Scales to Measure the Temperature
By combining two previously mentioned gas laws, we get:
for a fixed quantity of gas. This relationship is known as the ideal gas law. In a more generalized form, it is expressed as:
where:
This expression is referred to as the ideal gas equation.
According to the ideal gas equation, This proportionality allows gases to be used in temperature measurement through volume gas thermometers, where p ∝ T when the gas volume is kept constant.
Boyle’s Law states that ‘The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if both the temperature and amount of gas remain unchanged’. In mathematical terms this law is given as:
P ∝ 1/V or that PV = K
where P=Pressure of the gas; V=Volume of the gas; K=constant. It means that both the pressure and volume of a given mass of gas are inversely proportional to each other at a constant temperature. Furthermore, it also expresses that the product of pressure and volume for any gas is a constant and thus can be used to study the comparison of the gas under different conditions as:
P’V’ = P”V”
where both the products are for the same gas but under different pressures and volumes.
Charles’ law states that ‘ When the pressure of a sample of air is held constant, then the volume of the gas is directly proportional to its temperature‘, that is
V ∝ T
where V= Volume of a gas sample; T= Absolute temperature. Quite simply put, it says that Gases expand on heating and contract on cooling.
Avogadro’s law states that ‘Equal volumes of all gases at conditions of same temperature and pressure have the same number of molecules’, written as:
V ∝ n or V/n =K
where V=volume of gas; n = Number of moles (1 mole=6.022 x 10^{23} molecules). It implies that under similar conditions of pressure, volume and temperatures all gases will have an equal number of molecules, independent of the weight and density of the gas.
If we combine the results of all the above gas laws we get an equation that holds true for an ideal gas. The most common form of this equation is since PV= K and V/T =k then
PV/T = constant
Thus, the Ideal Gas Equation is given as
PV = n R T
where P= pressure of the gas; V=volume of the gas; n = Number of Moles; T = Absolute temperature; R = Ideal Gas constant also known as Boltzmann Constant = 0.082057 L atm K^{1} mol^{1}.
Using this equation, the study of any gas is possible under assumptions of STP conditions and subjecting the gas to reasonable restrictions to make it behave similarly to an Ideal gas.
What is an STP condition?
STP is short for ‘Standard Temperature and Pressure’. STP condition is defined (as per the International Standard Metric Conditions) as the surrounding absolute temperature of 288.15 Kelvin (15° Celsius) and a pressure of 1 atmosphere i.e. 1 bar or 101.325 K pa.
Thermodynamic temperature, which is also known as ‘Absolute Temperature’ is a basic parameter for the study of thermal properties of matter.
Using an ideal gas equation we can use gas for measuring temperature accurately. Again, as this law is universal in nature, all gases can be used to get the accurate temperature irrespective of their masses and other physical properties.
However, in actual use, it is observed that real gases often deviate from this law as compared to an ideal gas. But over a wide temperature base, all real gases more or less follow a linear path as can be seen in the case of gases taken above with pressure on the Yaxis and temperature on the Xaxis.
Assuming that gases continue to be in a gaseous state at lower temperatures we can extrapolate the lowest minimum temperature for a gas. This lowest possible temperature is called Absolute Zero. The temperature is obtained at 273.15° Celsius on the Celsius scale.
It is also responsible for the creation of a new temperature scale called as ‘Kelvin’ scale where absolute zero is taken as 0 and so this scale is also called as Absolute temperature scale. It is represented as Kelvin. Thus,
T(Celsius) = T(Kelvin) + 273.15
This conversion helps to determine the absolute temperature of a gas under study. Thus, we have finally derived the Ideal Gas equation after studying all the concepts involved which make up the gas laws. Thorough knowledge of these concepts is paramount to ensure the proper understanding of properties of a gas being subjected to thermal study and thus its applications.
Q2: How many moles of ‘He’ are contained in a 6litre canister at 101 KPa and 27 ° C. Take R= 8.314 J/mol K
Ans: Using the Ideal gas equation, n = PV/RT
Therefore, on substituting the values(T = 27 + 273 = 300) we get,
= 101 x 6/ 8.314 x 300 = 606/2494.2 = 0.2429 moles
Hence, 0.2429 moles of ‘He’ are contained in a 6litre canister at 101 KPa and 27°C
Most substances expand with heat and contract with cold. This change in dimensions with temperature variation is known as thermal expansion.
Example of Thermal Expansion
Three types of expansion occur in solids:
The expansion in length of a body due to increase in its temperature is called the linear expansion.
Linear expansion is expressed by the formula:
Here, αl represents the coefficient of linear expansion for the specific solid.
The unit of α is per degree Celsius (°C^1) in the CGS system and per Kelvin (K^1) in the SI system.
Linear Expansion
It involves the increase in the surface area of a substance when heated. A slight change in temperature, ΔT, leads to deformation, where the fractional change in surface area, ΔA/A, is directly proportional to ΔT.
Area Expansion
Here, αA represents the coefficient of area expansion of the given solid.
It refers to the fractional change in the volume of a substance. A slight variation in temperature, ΔT, causes deformation, where the fractional change in volume, ΔV/V, is directly proportional to ΔT.
Volume Expansion
Here, α_{V} is another characteristic of the substance, but it varies with temperature.
The coefficient of volume expansion only becomes constant at high temperatures. For instance, ethyl alcohol has a higher coefficient of volume expansion than mercury and thus expands more for the same temperature rise.
The graph depicts the Coefficient of volume expansion of copper concerning temperature:
Coefficient of volume expansion of copper changes with temperature
Q3: Define the term thermal expansion.
Ans: Most substances expand with heat and contract when cooled, changing dimensions with temperature shifts.
Q4: Calculate the pressure needed to maintain the length of a steel wire when heated by 100°C.
Ans: Given:
ΔT = 100°C,
Y = 2 x 10^11 Nm^−2,
α = 1.1 x 10^−5 K^−1.
Thermal strain = 2.2 x 10^8 Pa.
Specific heat of water: s = 4200 J/kg°C = 1000 cal/kg°C = 1 Kcal/kg°C = 1 cal/gm°C.
Specific heat of steam = half of specific heat of water = specific heat of ice
The heat capacity of a body is defined as the amount of heat required to raise the temperature of that body by 1°C. If `m' is the mass and `s' the specific heat of the body, then
Heat capacity = m s
Units of heat capacity in CGS system is, cal °C^{1} ; SI unit is, JK^{1}
It is the amount of water which requires the same amount of heat for the same temperature rise as that of the object
ms AT = m_{w} S_{w} ΔT
In calorie s_{w} = 1
∴ m_{w} = ms
m_{w} is also represented by W
so W = ms
Calorimetry is the branch of science focused on measuring heat. When a hot body comes into contact with a cooler one, heat flows from the warmer to the cooler body until they reach the same temperature.
According to the principle of calorimetry, the heat lost by the hotter body equals the heat gained by the colder body.
For example, if there are two bodies with masses m_{1 }and m_{2}, and specific heats s_{1} and s_{2}, then for a temperature change ΔT:
A calorimeter is a tool for measuring heat. It consists of a copper container with a stirrer and is placed in a wooden jacket insulated with materials like glass wool. This setup prevents heat exchange with the surroundings. The polished surfaces further reduce heat loss by radiation.
Calorimeter
The calorimeter has holes for inserting a thermometer and a stirrer. When different temperature bodies are placed inside, heat is exchanged until thermal equilibrium is reached. If no heat is lost to the surroundings, the heat gained by colder bodies equals the heat lost by hotter bodies, following the principle of calorimetry.
The process of changing a substance from one state to another, known as a "change of state," typically involves the absorption or release of heat.
Matter generally exists in three main states:
These states can transition from one to another through processes like:
These transitions occur due to heat exchange between the substance and its environment.
Effect of heat on ice
Plot of change of state of ice on heating
Melting and Melting Point:
Fusion and Freezing Point:
Vaporisation and Boiling Point:
Sublimation:
Effect of Pressure on Boiling Point:
The amount of heat transferred per unit mass during the change of phase of a substance without any change in its temperature is called latent heat of the substance for particular change.
Latent heat is denoted by L and having SI unit J kg−1. The value of latent heat is usually quoted at standard atmospheric pressure because it also depends upon the pressure.
Thus, if a mass m of a substance undergoes a change from one state to the other, then the quantity of heat required is given by
Hence, during the phase change, the heat required by the substance depends on the mass $mm$m of the substance and heat of transformation $Q$Q.
(a) Latent heat of Fusion (L_{f})
The heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm. pressure is called latent heat of fusion.
SI unit is J/Kg.
(b) Latent heat of vaporisation (L_{v})
The heat supplied to a substance which changes it from liquid to vapour state at its boiling point and 1 atm. pressure is called latent heat of vaporization.
SI unit is J/Kg.
If in question latent heat of water are not mentioned and to solve the problem it require to assume that we should consider following values.
Latent heat of ice : L = 80 cal/gm = 80 Kcal/kg = 4200 × 80 J/kg
Latent heat of steam : L = 540 cal/gm = 540 Kcal/kg = 4200 × 540 J/kg
The given figure, represents the change of state by different lines,
Temperature versus hat for water at 1 atm pressure
Note : If we increases the temperature of liquid (phase) but at a later time K .E. stop increasing and the phase of the liquid starts changing.
Q5: Heat required to increases the temperature of 1 kg water by 20°C
Ans: Heat required = ΔQ = m s Δθ
= 1 x 20 = 20 Kcal.
∴ S = 1 cal/gm°C = 1 Kcal/kg°C
Important Points:
(a) We know, s = , if the substance undergoes the change of state which occurs at constant temperature (DT = 0), the s = Q/0 = ¥. Thus the specific heat of a substance when it melts or boils at constant temperature is infinite.
(b) If the temperature of the substance changes without the transfer of heat (Q = 0) then s = = 0. Thus when liquid in the thermos flask is shaken, its temperature increases without the transfer of heat and hence and the specific heat of liquid in the thermos flask is zero.
(c) To raise the temperature of saturated water vapour, heat (Q) is withdrawn. Hence, specific heat of saturated water vapour is negative. (This is for your information only and not in the course)
(d) The slight variation of specific heat of water with temperature is shown in the graph at 1 atmosphere pressure. Its variation is less than 1% over the interval form 0 to 100°C.
Q6: An iron block of mass 2 kg, fall from a height 10 m. After colliding with the ground it loses 25 % energy to surroundings. Then find the temperature rise of the block (Take specific heat of iron 470 J/kg° C)
Ans:
Q7: The temperature of equal masses of three different liquids A, B, and C are 10°C 15°C and 20°C respectively. The temperature when A and B are mixed is 13°C and when B and C are mixed, it is 16°C. What will be the temperature when A and C are mixed?
Ans:
when A and B are mixed
mS_{1} x (13  10) = m x S_{2} x (15  13)
3S_{1} = 2S_{2} ...(1)
when B and C are mixed
S_{2} x 1 = S_{3} x 4 ...(2)
when C and A are mixed
S_{1}(θ  10) = s_{3} x (20  θ) ,..(3)
by using equation (1), (2) and (3)
Q8: 500 gm of water at 80°C is mixed with 100 gm steam at 120°C. Find out the final mixture.
Ans: 120°C steam 100°C steam
Req. heat = 100 × ×20 = 1 kcal
80°C water 100°C water
Req. heat = 500 × 1 × 20 = 10 kcal
100gm steam 100 gm water at 100°C
Req. heat = 100 × 540 = 54 kcal
Total heat = 55 kcal.
Remaining heat = 55  10 = 45 kcal
Now we have 600 gm water at 100°C ⇒ 4500 = m × 540 ⇒
So at last we have gm steam and of water
102 videos411 docs121 tests

1. What is the Ideal Gas Equation and how is it used in thermodynamics? 
2. How does thermal expansion occur and what are its implications in everyday life? 
3. Why is water considered unique in its thermal properties compared to other substances? 
4. What is Young's Modulus and how does it relate to the stressstrain relationship in materials? 
5. What is the Law of Mixtures in relation to thermal properties, and how is it applied? 

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