18Calorimetry, Change of State and Hygrometry

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CALORIMETRY, CHANGE OF STATE AND HYGROMETRY
▢ Calorimetry:
❖ Principle: Heat lost by one body = Heat gained by another body
❖ Based On: Conservation of energy

Table 1: Calorimetry Basic Formulae

Quantity
Formula
Heat lost/gained
\(\Delta Q=mc\Delta\theta\)
Thermal capacity
\(T.C.=\frac{\Delta Q}{\Delta\theta}\)
Thermal capacity also
\(T.C.=mc=\mu C\)
Specific heat capacity
\(c=\frac{\Delta Q}{m\Delta\theta}\)
▢ Thermal Capacity:
❖ Definition: Amount of heat lost or gained by a body to change its temperature by 1°C or 1K

Table 1: Thermal Capacity

Point
Value / Formula
Formula
\(T.C.=mc=\mu C\)
Also
\(T.C.=\frac{\Delta Q}{\Delta\theta}\)
\(m\)
Mass
\(c\)
Specific heat capacity
\(\mu\)
Number of moles
\(C\)
Molar heat capacity
Unit
\(cal/^{\circ}C\) or \(J/K\)
Dimension
\([ML^2T^{-2}K^{-1}]\)
▢ Specific Heat Capacity:
❖ Definition: Amount of heat required to change temperature of unit mass of a substance by 1°C or 1K

Table 1: Specific Heat Capacity

Point
Value / Formula
Formula
\(c=\frac{\Delta Q}{m\Delta\theta}\)
Unit
\(cal\ g^{-1}\ ^\circ C^{-1}\) or \(J\ kg^{-1}K^{-1}\)
Dimension
\([L^2T^{-2}K^{-1}]\)
Depends on
State of substance
Order
Solids < Liquids < Gases
❖ Dulong-Petit Law:
◉ Formula: \(c\times Atomic\ weight=6.4\)
◉ Relation: \(Specific\ heat\propto\frac{1}{Atomic\ weight}\)
◉ Conclusion: Heavier element → smaller specific heat
❖ Important Values:

Table 1: Specific Heat Values

Substance / Condition
Specific heat
Ice
\(0.5\ cal\ g^{-1}\ ^\circ C^{-1}\)
Water
\(1\ cal\ g^{-1}\ ^\circ C^{-1}\)
Hydrogen
\(3.5\ cal\ g^{-1}\ ^\circ C^{-1}\)
Radon and Actinium
\(0.22\ cal\ g^{-1}\ ^\circ C^{-1}\)
Saturated water vapour
Negative
Isothermal condition
\(\infty\)
Adiabatic condition
0
❖ Special Points:
  • •
    Water has highest specific heat among common solids and liquids
  • •
    Water is used as coolant due to high specific heat
  • •
    During change of state, specific heat is infinite
  • •
    Specific heat inside thermos under adiabatic condition is zero
▢ Water Equivalent:
❖ Definition: Mass of water which absorbs or gives same heat as the body for same temperature rise

Table 1: Water Equivalent

Quantity
Formula / Point
Water equivalent
\(W=\frac{mc}{c_w}\)
In CGS
\(c_w=1\ cal\ g^{-1}\ ^\circ C^{-1}\)
In CGS
\(W=mc\)
Unit
kg or g
Relation
Water equivalent in gram = thermal capacity in \(cal/^{\circ}C\)
In CGS \(c_w=1\)
▢ Molar Heat Capacity:
❖ Definition: Heat required per mole to increase temperature by unity

Table 1: Molar Heat Capacities

Quantity
Formula / Value
At constant volume
\(C_v=\frac{\Delta Q}{\mu\Delta T}\)
At constant pressure
\(C_p=\frac{\Delta Q}{\mu\Delta T}\)
Number of moles
\(\mu=\frac{m}{M}\)
Ratio of specific heats
\(\gamma=\frac{C_p}{C_v}\)
Monoatomic gas
\(\gamma=1.67\)
Diatomic gas
\(\gamma=1.40\)
Polyatomic gas
\(\gamma=1.33\)
Meyer's relation
\(C_p-C_v=R\)
❖ Why \(C_p>C_v\): At constant pressure, gas expands and work is done against external pressure
▢ Latent Heat:
❖ Definition: Heat required to change state of unit mass of substance without change in temperature

Table 1: Latent Heat Basics

Point
Value / Formula
Formula
\(Q=mL\)
Unit
\(cal/g\) or \(J/kg\)
Dimension
\([L^2T^{-2}]\)
Temperature during change of state
Constant
Nature of phase transformation
Isothermal change
❖ Types:

Table 1: Latent Heat Types

Type
Meaning
Value
Latent heat of fusion
Heat absorbed in solid → liquid or liberated in liquid → solid
Ice: \(80\ cal/g=3.36\times10^5\ J/kg=6\times10^4\ J/mol\)
Latent heat of vaporization
Heat absorbed in liquid → gas or liberated in gas → liquid
Water: \(540\ cal/g=2.26\times10^6\ J/kg=4.08\times10^4\ J/mol\)
❖ Change of State Energy:
  • •
    Heat supplied increases internal potential energy
  • •
    Heat supplied does external work against pressure
  • •
    Internal kinetic energy remains constant
  • •
    Steam at 100°C has greater internal energy than water at 100°C
  • •
    If surrounding pressure increases, latent heat of steam decreases
  • •
    Molecules moving apart → energy absorbed
  • •
    Molecules coming close → energy released
▢ Hoar Frost:
❖ Definition: Direct conversion of vapour into solid
❖ Nature: Converse of sublimation
❖ Example: Formation of snow by freezing of cloud
▢ Effect of Pressure on Melting Point:
❖ Clausius-Clapeyron: Effect of pressure on melting point depends on change in volume during melting

Table 1: Pressure and Melting Point

Condition
Effect
Examples
\(V_2>V_1\)
Melting point increases with pressure
Sulphur, glass, ghee, gold, wax
\(V_2
Melting point decreases with pressure
Ice, rubber
❖ Special Points:
  • •
    Melting point decreases on adding soluble impurities
  • •
    Melting point of ice decreases by 1°C with pressure increase of nearly 133 atm
  • •
    Regelation = melting of ice due to pressure and resolidification after removal of pressure
▢ Effect of Pressure on Boiling Point:
❖ Rule: With increase in pressure, boiling point of all liquids increases

Table 1: Boiling Point Points

Condition
Effect
Pressure increases
Boiling point increases
Soluble impurity added
Boiling point increases
Pressure decreases by 5 mmHg
Boiling point of water decreases by 1°C
At high altitude
Pressure decreases, so boiling point decreases
Pressure cooker
Pressure increases, boiling point increases, cooking faster
❖ Boiling Condition: Water boils when saturated vapour pressure becomes equal to atmospheric pressure
▢ Hygrometry:
❖ Definition: Measurement of amount of water vapour present in atmosphere

Table 1: Hygrometry Terms

Term
Meaning
Saturated vapour
Air containing maximum possible amount of water vapour
Saturated vapour pressure / SVP
Pressure of water vapour in saturated air
SVP at 0°C
4.6 mmHg
Unsaturated air
Air which can accommodate more water vapour
Unsaturated vapour pressure / UVP
Actual vapour pressure less than SVP
▢ SVP and UVP:

Table 1: Properties of SVP and UVP

Property
Point
UVP vs SVP
UVP is always less than SVP
SVP depends on
Nature of liquid and temperature
Temperature increases
SVP increases
SVP independent of
Volume occupied by vapour
SVP independent of
Other vapours present
SVP and gas laws
SVP does not obey gas laws
UVP and gas laws
UVP obeys gas laws
Total vapour pressure
\(P=P_1+P_2+P_3+...\)
▢ Relative Humidity:
❖ Definition 1: Ratio of actual mass of water vapour present to mass required to saturate same volume of air at same temperature
❖ Definition 2: Ratio of actual vapour pressure to saturated vapour pressure at same temperature

Table 1: Relative Humidity Formulae

Formula
Meaning
\(R.H.=\frac{m}{M}\times100\%\)
Mass form
\(R.H.=\frac{p}{P}\times100\%\)
Pressure form
\(R.H.=\frac{SVP\ at\ dew\ point}{SVP\ at\ room\ temperature}\times100\%\)
Dew point form
❖ Important Points:
  • •
    R.H. is low in dry air
  • •
    R.H. is high in moist air
  • •
    Comfortable R.H. for humans = 60%–65%
  • •
    If atmospheric temperature and dew point are nearly equal, R.H. is nearly 100%
▢ Dew Point:
❖ Definition: Temperature at which water vapour actually present in atmosphere is just sufficient to saturate it
❖ Condition: At dew point, actual vapour pressure at room temperature = saturated vapour pressure
❖ Special Points:
  • •
    In absolutely dry air, no dew point is observed
  • •
    Dew formation occurs in early morning due to condensation of saturated vapour
  • •
    Sprinkling water in room increases both R.H. and dew point
▢ Absolute Humidity:
❖ Definition: Amount of water vapour actually present in unit volume of atmosphere
▢ Triple Point:
❖ Definition: Point at which solid, liquid and vapour states coexist in equilibrium simultaneously

Table 1: Triple Point of Water

Quantity
Value
Pressure
4.58 mmHg
Temperature
0.01°C = 273.16 K
▢ Wet Bulb and Dry Bulb Hygrometer:

Table 1: Principle

Condition
Result
Difference in readings
Related to relative humidity
Higher R.H.
Less difference between wet and dry bulb readings
R.H. = 100%
Wet bulb reading = dry bulb reading
R.H. = 100% and room temperature = \(\theta^\circ C\)
Dew point = \(\theta^\circ C\)
▢ Important Calorimetry Results:

Table 1: Ice-Water-Steam Mixing

Case
Result
1 g steam at 100°C vs 1 g water at 100°C
Steam causes more severe burn due to latent heat of vaporization
Equal masses of ice and steam mixed
Final temperature = 100°C
Steam at 100°C required to melt \(m\) g ice at 0°C
\(\frac{m}{8}\) g
Equal mass ice at 0°C + water at \(\theta^\circ C\), \(\theta\leq80^\circ C\)
Final temperature = 0°C
Water formed for \(\theta\leq80^\circ C\)
\(\frac{m}{80}(80+\theta)\)
Ice remaining for \(\theta\leq80^\circ C\)
\(\frac{m}{80}(80-\theta)\)
Water : ice ratio for \(\theta\leq80^\circ C\)
\(\frac{80+\theta}{80-\theta}\)
Equal mass ice at 0°C + water at \(\theta^\circ C\), \(\theta>80^\circ C\)
All ice melts
Final temperature for \(\theta>80^\circ C\)
\(\frac{\theta-80}{2}\)
Amount of water for \(\theta>80^\circ C\)
\(2m\)
▢ Read and Digest:

Table 1: Important Points

Fact
Point
Steam burn
More severe due to latent heat of vaporization
Pressure around steam increases
Latent heat of steam decreases
Heat during melting
Used to increase average intermolecular distance
Ice melting point
Decreases with pressure
Specific heat range
0 to \(\infty\)
Adiabatic specific heat
0
Isothermal specific heat
\(\infty\)
Iceberg melts at base
High pressure lowers melting point
Pressure cooker
Boiling point of water increases
Water under vacuum
Evaporation rate increases
Evaporation rate
Increases with temperature and decreases with external pressure
Evaporation effect
Remaining liquid cools
Water at 0°C in open container placed in vacuum
Part vaporizes and rest freezes
Calorific value of fuel
Determined by bomb calorimeter
More volatile substance
Lower boiling point
Water in car radiator
Due to high specific heat
High altitude cooking
Takes longer due to lower boiling point
Sprinkling water in closed room
Reduces temperature due to high latent heat of vaporization
Boiling water extinguishes fire quickly
Due to high heat content and steam formation
Deep mine water boiling
Temperature > 100°C due to high pressure
\(C_p
True only for material which contracts on heating
Specific heat in °C vs °F
Numerical value is greater in centigrade scale
Cooking utensil
Low specific heat and high thermal conductivity
Dry air
Low R.H.
Moist air
High R.H.
Absolutely dry air
No dew point
Comfortable R.H.
60%–65%
▢ High-Yield Recall:

Table 1: Calorimetry, Change of State and Hygrometry One-Liners

Fact
Answer
Calorimetry principle
Heat lost = Heat gained
Calorimetry based on
Conservation of energy
Thermal capacity
\(T.C.=mc=\mu C\)
Specific heat
\(c=\frac{\Delta Q}{m\Delta\theta}\)
Heat equation
\(\Delta Q=mc\Delta\theta\)
Water equivalent
\(W=\frac{mc}{c_w}\)
In CGS water equivalent
\(W=mc\)
Molar heat at constant volume
\(C_v=\frac{\Delta Q}{\mu\Delta T}\)
Molar heat at constant pressure
\(C_p=\frac{\Delta Q}{\mu\Delta T}\)
Meyer's relation
\(C_p-C_v=R\)
Gamma
\(\gamma=\frac{C_p}{C_v}\)
Monoatomic gas gamma
1.67
Diatomic gas gamma
1.40
Polyatomic gas gamma
1.33
Latent heat
\(Q=mL\)
Latent heat of fusion of ice
\(80\ cal/g\)
Latent heat of vaporization of water
\(540\ cal/g\)
Hoar frost
Vapour → solid
Ice melting point with pressure
Decreases
Boiling point with pressure
Increases
Regelation
Melting by pressure and freezing after pressure removal
Hygrometry
Measurement of water vapour in air
SVP at 0°C
4.6 mmHg
R.H. mass formula
\(R.H.=\frac{m}{M}\times100\%\)
R.H. pressure formula
\(R.H.=\frac{p}{P}\times100\%\)
Dew point
Temperature at which air becomes just saturated
Absolute humidity
Water vapour per unit volume of air
Triple point of water
0.01°C, 4.58 mmHg
Wet bulb-dry bulb high R.H.
Small reading difference
R.H. 100%
Dew point = room temperature
Q1.
Heat required to convert 1 gm of ice at 0°C to steam at 100°C;
📅BP 2012
Q2.
Water is used as coolant due to:
📅BP 2011
Q3.
As compared to a person with white skin, another person with dark skin will experience
📅BP 2011
Q4.
When ice cube is placed on a table and melts, which is correct?
📅IOM 2012
Q5.
A body of mass 100 gm was given a heat of 420 J. Find the raise in temperature? (specific heat capacity = 420 J/kg·K)
📅MOE 2010
Q6.
A 10 kg iron bar (specific heat 0.11 cal/gm°C) at 80°C is placed on ice (Lf=80 cal/gm). How much ice melts?
📅MOE 2010
Q7.
Amount of heat required to change 2 kg water from 20°C to 40°C.
📅MOE 2011
Q8.
When relative humidity is 100%, the room temperature is equal to:
📅MOE 2011-2013
Q9.
Final temperature when mixing 0.5 kg ice at 0°C with 0.5 kg water at 75°C:
📅MOE 2011
Q10.
Temperature at which water vapor in atmosphere is saturated:
📅IOM
Q11.
The door of a running refrigerator is opened. Which is true?
📅IE 2011
Q12.
Steam at 100°C is poured into 1.1kg water at 15°C (calorimeter water equivalent=0.02kg). Mass of steam condensed if final temp=80°C?
📅IE 2012
Q13.
Three liquids A(10°C), B(25°C), C(40°C). When A+B mix, temp=15°C; B+C mix, temp=30°C. What is A+C mixture temp?
📅IE 2013
Q14.
Heat required to convert 1gm ice at -10°C to steam at 100°C?
📅KU 2010
Q15.
Liquids with volume ratio 1:1, density ratio 1:3, specific heat ratio 3:1. Their heat capacity ratio?
📅MOE 2013
Q16.
Substance that expands on both heating and cooling:
📅BP 2012
Q17.
5kg mass needs 80J heat for 10K rise. Specific heat capacity in Jkg-1K-1?
📅IOM 2014
Q18.
Density of ice is:
📅KU 2010
Q19.
When relative humidity=100% at 30°C, dew point is:
📅BP 2014
Q20.
50gm ice at 0°C + 50gm water at 80°C. Final temperature?
📅MOE 2009
Q21.
10kg iron bar (c=0.11 cal/gm°C) at 80°C placed on ice (Lf=80 cal/gm). Ice melted?
📅MOE 2010
Q22.
Heat to convert 1gm ice at -100°C to steam at 100°C?
📅OM 2001
Q23.
10gm ice at -10°C → steam at 100°C. Heat required?
📅MOE 2006
Q24.
Melting point of ice:
📅MOE 2005
Q25.
Heat to convert 1gm ice at 0°C → steam at 100°C?
📅MOE 2003
Q26.
Heat required to melt 1gm ice without temp change:
📅MOE 2002
Q27.
Energy to change 1kg ice from -10°C to 50°C (cice=0.5 cal/gm°C, Lf=80 cal/gm)?
📅MOE 2006
Q28.
When ice melts:
📅MOE
Q29.
When liquid changes to vapor, increasing pressure causes boiling point to:
📅MOE
Q30.
When two ice blocks are pressed together, they join because:
Q31.
Steam at 100°C passed into 1.1kg water + 0.02kg calorimeter at 15°C → 80°C. Mass of steam condensed?
📅TE-2007
Q32.
1gm ice at 0°C + 1gm steam at 100°C mixed. Resulting temp?
Q33.
25g water at 46°C + 10g ice at 0°C. Resulting temp?
Q34.
20g ice at 0°C + 20g water at 60°C. Final water mass?
Q35.
Equal masses: ice at -10°C + water at 60°C. How much ice melts?
Q36.
Specific heats C1 (cal/gm°C) and C2 (cal/gm°F). Valid relation?
Q37.
Density ratio 3:4, specific heat ratio 4:3. Thermal capacity per unit volume ratio?
📅IOM
Q38.
Sphere radii ratio 4:9, specific heat ratio 9:4. Thermal capacity ratio?
Q39.
50g copper at 100°C placed on ice. Ice melted? (cCu=0.1 cal/gm°C, Lf=80 cal/gm)
Q40.
Cooking is fast in pressure cooker because:
📅MOE/KU
Q41.
Water falls 84m. Half KE → heat. Temperature rise? (g=10m/s²)
Q42.
Liquids at 20°C and 40°C. Same mass mixed → 32°C. Specific heat ratio?
Q43.
Steam at 100°C → 1.1kg water + 0.02kg calorimeter at 15°C → 80°C. Steam condensed?
📅IE 2007
Q44.
10g ice at 0°C + 55g water equivalent tumbler at 40°C. Final temp? (L=80 cal/g)
Q45.
Water at -10°C in insulated container + ice crystal. Ratio of ice formed to initial water?
Q46.
Heat from condensing x g steam at 100°C converts y g ice at 0°C → water at 100°C. x:y?
Q47.
100% RH at 30°C → dew point is:
Q48.
Man feels hottest when relative humidity is:
Q49.
Dew formation at 4.6°C, dew at 5.4°C, air temp=20°C. RH? (SVP at 5°C=6.5mmHg, 20°C=17.5mmHg)
📅IOM 2010
Q50.
Geyser ejects 1L/min (22°C → 37°C). Power?
📅IOM 2016
Q51.
Man chews 60gm ice/min (Lf=80 cal/gm). Power?
📅IOM 2017
Q52.
When water is heated steadily, temperature stops rising when it starts to:
📅KU 2017