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.
If 1 g steam at 100°C is mixed with 1 g ice at 0°C, resultant temperature is
📅CEE MBBS 2025
Q2.
Water has temperature 22°C. A geyser ejects 1 L water per minute at 37°C. What is the power of geyser?
📅CEE Nursing 2024
Q3.
What heat is required to raise 1 g of water from 14.5°C to 15.5°C?
📅CEE MBBS 2023
Q4.
The ratio of masses of two substances is 3:2 and the ratio of their specific heats is 1:4. If equal amounts of heat are supplied, the ratio of rise in temperature is:
📅CEE MBBS 2023
Q5.
The saturation vapour pressure of water at 100°C is:
📅CEE MBBS 2023
Q6.
If 420 J is lost by 20 g of water, what is the change in temperature?
📅CEE MBBS 2021
Q7.
Equal masses of two liquids having specific heat capacities S₁ at 20°C and S₂ at 40°C are mixed to obtain a final temperature of 32°C. Then S₁:S₂ is:
📅CEE MBBS 2021
Q8.
Calculate the amount of energy required to melt 1 kg of ice at 0°C, given heat of fusion of water is 334 J/gm.
📅CEE BDS 2023
Q9.
How much heat is required to convert \(1\,\mathrm g\) of ice at \(0^\circ\mathrm C\) into steam at \(100^\circ\mathrm C\)? Use \(L_f=80\,\mathrm{cal/g}\), \(c_w=1\,\mathrm{cal/(g\,K)}\), and \(L_v=540\,\mathrm{cal/g}\).
📅BP 2012
Q10.
Water is used as a coolant primarily because it has:
📅BP 2011
Q11.
Compared with an otherwise identical light-coloured surface, an ideal dark surface under the same conditions generally:
📅BP 2011
Q12.
As an ice cube melts at its melting point, what happens to the average potential energy of its molecules?
📅IOM 2012
Q13.
A body of mass \(100\,\mathrm g\) receives \(420\,\mathrm J\) of heat. If its specific heat capacity is \(420\,\mathrm{J/(kg\,K)}\), what is its temperature rise?
📅MOE 2010
Q14.
An iron bar of mass \(10\,\mathrm{kg}\), initially at \(80^\circ\mathrm C\), is placed on sufficient ice at \(0^\circ\mathrm C\). How much ice melts? Take \(c_{\mathrm{Fe}}=0.11\,\mathrm{cal/(g\,K)}\) and \(L_f=80\,\mathrm{cal/g}\).
📅MOE 2010
Q15.
How much heat is required to warm \(2\,\mathrm{kg}\) of water from \(20^\circ\mathrm C\) to \(40^\circ\mathrm C\)? Take \(c_w=4200\,\mathrm{J/(kg\,K)}\).
📅MOE 2011
Q16.
When relative humidity is 100%, the air temperature is equal to its:
📅MOE 2011-2013
Q17.
\(0.5\,\mathrm{kg}\) of ice at \(0^\circ\mathrm C\) is mixed with \(0.5\,\mathrm{kg}\) of water at \(75^\circ\mathrm C\) in an insulated vessel. What is the equilibrium temperature? Take \(L_f=80\,\mathrm{cal/g}\) and \(c_w=1\,\mathrm{cal/(g\,K)}\).
📅MOE 2011
Q18.
The temperature to which moist air must be cooled, without changing its water-vapour content, to become saturated is called the:
📅IOM
Q19.
The door of an operating refrigerator is left open in an otherwise closed, insulated room. What happens to the room temperature?
📅IE 2011
Q20.
Steam at \(100^\circ\mathrm C\) is introduced into \(1.1\,\mathrm{kg}\) of water at \(15^\circ\mathrm C\). The calorimeter has water equivalent \(0.02\,\mathrm{kg}\). If the final temperature is \(80^\circ\mathrm C\), what mass of steam condenses? Use \(L_v=2.268\times10^6\,\mathrm{J/kg}\) and \(c_w=4200\,\mathrm{J/(kg\,K)}\).
📅IE 2012
Q21.
Three liquids A, B, C are at \(10^\circ\mathrm C\), \(25^\circ\mathrm C\), and \(40^\circ\mathrm C\). Mixing A and B gives \(15^\circ\mathrm C\), and mixing B and C gives \(30^\circ\mathrm C\). With the same original portions and no heat loss, find the temperature upon mixing A and C.
📅IE 2013
Q22.
Find the heat needed to change \(1\,\mathrm g\) of ice at \(-10^\circ\mathrm C\) to steam at \(100^\circ\mathrm C\). Use \(c_i=0.5\), \(c_w=1\), \(L_f=80\), and \(L_v=540\) in calorie–gram units.
📅KU 2010
Q23.
Two liquids have volumes in the ratio 1:1, densities in the ratio 1:3, and specific heat capacities in the ratio 3:1. Find their total heat-capacity ratio.
📅MOE 2013
Q24.
Which substance expands when its temperature is changed slightly in either direction from the stated temperature?
📅BP 2012
Q25.
If \(5\,\mathrm{kg}\) of a substance needs \(80\,\mathrm J\) for a temperature rise of \(10\,\mathrm K\), what is its specific heat capacity in \(\mathrm{J/(kg\,K)}\)?
📅IOM 2014
Q26.
Compared with liquid water near its freezing point, the density of ice is:
📅KU 2010
Q27.
If relative humidity is \(100\%\) at \(30^\circ\mathrm C\), what is the dew-point temperature?
📅BP 2014
Q28.
\(50\,\mathrm g\) of ice at \(0^\circ\mathrm C\) is mixed with \(50\,\mathrm g\) of water at \(80^\circ\mathrm C\). Find the equilibrium temperature. Take \(L_f=80\,\mathrm{cal/g}\).
📅MOE 2009
Q29.
An iron bar of mass \(10\,\mathrm{kg}\) at \(80^\circ\mathrm C\) is placed on ice at \(0^\circ\mathrm C\). Find the mass of ice melted if \(c_{\mathrm{Fe}}=0.11\,\mathrm{cal/(g\,K)}\) and \(L_f=80\,\mathrm{cal/g}\).
📅MOE 2010
Q30.
How much heat converts \(1\,\mathrm g\) of ice at \(-100^\circ\mathrm C\) into steam at \(100^\circ\mathrm C\)? Use \(c_i=0.5\), \(c_w=1\), \(L_f=80\), and \(L_v=540\) in calorie–gram units.
📅OM 2001
Q31.
How much heat is required to convert \(10\,\mathrm g\) of ice at \(-10^\circ\mathrm C\) into steam at \(100^\circ\mathrm C\)? Use \(c_i=0.5\), \(c_w=1\), \(L_f=80\), and \(L_v=540\) in calorie–gram units.
📅MOE 2006
Q32.
For ordinary ice near its melting point, an increase in external pressure causes its melting point to:
📅MOE 2005
Q33.
How much heat is required to convert \(1\,\mathrm g\) of ice at \(0^\circ\mathrm C\) into steam at \(100^\circ\mathrm C\), taking \(L_f=80\), \(c_w=1\), and \(L_v=540\) in calorie–gram units?
📅MOE 2003
Q34.
How much heat is absorbed when \(1\,\mathrm g\) of ice melts at \(0^\circ\mathrm C\) without a temperature change?
📅MOE 2002
Q35.
Calculate the heat needed to convert \(1\,\mathrm{kg}\) of ice at \(-10^\circ\mathrm C\) into water at \(50^\circ\mathrm C\). Take \(c_i=0.5\), \(c_w=1\), and \(L_f=80\) in calorie–gram units.
📅MOE 2006
Q36.
During melting at its melting point, ice:
📅MOE
Q37.
For an ordinary liquid, increasing the external pressure generally makes its boiling point:
📅MOE
Q38.
The textbook phenomenon of regelation, in which two pieces of ice can join when pressed together and released, is explained by:
Q39.
Steam at \(100^\circ\mathrm C\) enters \(1.1\,\mathrm{kg}\) of water in a calorimeter (water equivalent \(0.02\,\mathrm{kg}\)) at \(15^\circ\mathrm C\). The final temperature is \(80^\circ\mathrm C\). Calculate the mass condensed, with \(L_v=2.268\times10^6\,\mathrm{J/kg}\), \(c_w=4200\,\mathrm{J/(kg\,K)}\).
📅TE-2007
Q40.
\(1\,\mathrm g\) of ice at \(0^\circ\mathrm C\) is mixed with \(1\,\mathrm g\) of steam at \(100^\circ\mathrm C\) in an insulated vessel. What is the final equilibrium temperature at atmospheric pressure?
Q41.
\(25\,\mathrm g\) of water at \(46^\circ\mathrm C\) is mixed with \(10\,\mathrm g\) of ice at \(0^\circ\mathrm C\). Find the final temperature; use \(L_f=80\,\mathrm{cal/g}\) and \(c_w=1\,\mathrm{cal/(g\,K)}\).
Q42.
\(20\,\mathrm g\) of ice at \(0^\circ\mathrm C\) is mixed with \(20\,\mathrm g\) of water at \(60^\circ\mathrm C\). How much liquid water is present at equilibrium? Take \(L_f=80\,\mathrm{cal/g}\).
Q43.
Equal masses of ice at \(-10^\circ\mathrm C\) and water at \(60^\circ\mathrm C\) are mixed in an insulated vessel. What fraction of the initial ice melts? Use \(c_i=0.5\), \(c_w=1\), and \(L_f=80\) in calorie–gram units.
Q44.
Let \(c_1\) denote specific heat capacity in \(\mathrm{cal/(g\,^\circ C)}\) and \(c_2\) that in \(\mathrm{cal/(g\,^\circ F)}\). Which relation holds for the same material?
Q45.
Two liquids have density ratio 3:4 and specific heat ratio 4:3. Find their heat-capacity-per-unit-volume ratio.
📅IOM
Q46.
Two solid spheres have radii in the ratio 4:9 and specific heats in the ratio 9:4. Assuming equal densities, find their heat-capacity ratio.
Q47.
A \(50\,\mathrm g\) copper block at \(100^\circ\mathrm C\) is placed in excess ice at \(0^\circ\mathrm C\). Find the mass melted, given \(c_{\mathrm{Cu}}=0.1\,\mathrm{cal/(g\,K)}\) and \(L_f=80\,\mathrm{cal/g}\).
Q48.
Why does food cook faster in a pressure cooker?
📅MOE/KU
Q49.
Water falls through a height of \(84\,\mathrm m\), and half its lost gravitational potential energy becomes heat within the water. Find its temperature rise if \(g=10\,\mathrm{m/s^2}\) and \(c_w=4200\,\mathrm{J/(kg\,K)}\).
Q50.
Equal masses of two liquids at \(20^\circ\mathrm C\) and \(40^\circ\mathrm C\) are mixed. The final temperature is \(32^\circ\mathrm C\). Find the ratio of their specific heats \(c_1:c_2\) (cooler liquid first).
Q51.
Steam at \(100^\circ\mathrm C\) is passed into \(1.1\,\mathrm{kg}\) of water with calorimeter water equivalent \(0.02\,\mathrm{kg}\), initially at \(15^\circ\mathrm C\). If the final temperature is \(80^\circ\mathrm C\), find steam condensed. Take \(L_v=2.268\times10^6\,\mathrm{J/kg}\), \(c_w=4200\,\mathrm{J/(kg\,K)}\).
📅IE 2007
Q52.
\(10\,\mathrm g\) of ice at \(0^\circ\mathrm C\) is put into a tumbler initially at \(40^\circ\mathrm C\), whose water equivalent is \(55\,\mathrm g\). Find the final temperature, ignoring heat loss. Take \(L_f=80\,\mathrm{cal/g}\).
Q53.
Supercooled liquid water at \(-10^\circ\mathrm C\) is in a thermally insulated container. A tiny ice crystal triggers freezing and the final mixture reaches \(0^\circ\mathrm C\). What fraction of the initial water mass freezes? Take \(c_w=1\,\mathrm{cal/(g\,K)}\), \(L_f=80\,\mathrm{cal/g}\).
Q54.
Condensation of \(x\,\mathrm g\) of steam at \(100^\circ\mathrm C\) supplies heat to convert \(y\,\mathrm g\) of ice at \(0^\circ\mathrm C\) to water at \(100^\circ\mathrm C\). Find \(x:y\), taking \(L_v=540\), \(L_f=80\), and \(c_w=1\) in calorie–gram units.
Q55.
Air has relative humidity \(100\%\) at \(30^\circ\mathrm C\). Find its dew point.
Q56.
Under otherwise similar hot-weather conditions, a person tends to feel hottest when the relative humidity is:
Q57.
The dew-point temperature is about \(5^\circ\mathrm C\) while the air temperature is \(20^\circ\mathrm C\). Given saturation vapour pressures \(6.5\,\mathrm{mmHg}\) at \(5^\circ\mathrm C\) and \(17.5\,\mathrm{mmHg}\) at \(20^\circ\mathrm C\), find relative humidity.
📅IOM 2010
Q58.
A water heater supplies \(1\,\mathrm{L/min}\) of water, warming it from \(22^\circ\mathrm C\) to \(37^\circ\mathrm C\). Find heating power if \(c_w=4200\,\mathrm{J/(kg\,K)}\) and water density is \(1\,\mathrm{kg/L}\).
📅IOM 2016
Q59.
A person melts \(60\,\mathrm g\) of ice per minute by supplying heat at the melting point. What is the corresponding heating power if \(L_f=80\,\mathrm{cal/g}\) and \(1\,\mathrm{cal}=4.2\,\mathrm J\)?
📅IOM 2017
Q60.
When water is heated steadily at atmospheric pressure, its temperature stops rising at the boiling point because it begins to:
📅KU 2017