This chapter delves into the effects of heat on matter. Here’s a summary:
Thermal Expansion
Types:
Linear expansion (change in length)
Area expansion (change in area)
Volume expansion (change in volume)
Coefficient of expansion: A measure of how much a material expands with a change in temperature.
Calorimetry
Heat Capacity: The amount of heat required to raise the temperature of a body by 1 degree Celsius.
Specific Heat Capacity: The amount of heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius.
Latent Heat: The hidden energy used to change a substance’s form without changing its temperature.
Thermal Conductivity
Thermal Conductivity: The property of a material to conduct heat.
Thermal Resistance: The resistance offered by a material to the flow of heat.
Ideal Gas Equation
Equation of State: A mathematical expression that correlates the pressure, volume, and temperature of a gas.
Kinetic Theory of Gases: Explains the behavior of gases based on the motion of gas molecules.
Specific Heat of Gases
Specific Heat at Constant Pressure (Cp): Heat required to raise the temperature of a unit mass of gas by 1 degree Celsius at constant pressure.
Specific Heat at Constant Volume (Cv): Heat required to raise the temperature of a unit mass of gas by 1 degree Celsius at constant volume.
Newton’s Law of Cooling
The rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings.
This chapter provides a foundational understanding of how heat affects different materials and how various properties like expansion, heat capacity, and conductivity influence these effects.
1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Ans : Understanding the Problem:
We are given the temperatures of neon and carbon dioxide in Kelvin.
here, convert these temperatures to Celsius and Fahrenheit scales.
Converting to Celsius:
Given,
relationship between (K) and Celsius (°C);
°C = K – 273.15
For Neon:
T(K) = 24.57 K
T(°C) = 24.57 – 273.15 = -248.58°C
For CO2:
T(K) = 216.55 K
T(°C) = 216.55 – 273.15 = -56.60°C
Converting to Fahrenheit:
Given,
relationship between (K) and (°F)
°F = (9/5)(K – 273.15) + 32
For Neon:
T(K) = 24.57 K
T(°F) = (9/5)(24.57 – 273.15) + 32 = -415.44°F
For CO2:
T(K) = 216.55 K
T(°F) = (9/5)(216.55 – 273.15) + 32
= -69.88°F
Conclusion:
The temperature of neon is -248.58°C or -415.44°F.
Carbon dioxide has a temperature of -56.60 degrees Celsius or -69.88 degrees Fahrenheit.
2.Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?
Ans : *Understanding the Problem
*We’re given the triple point of water on two different absolute temperature scales, A and B. We need to find the relation between temperatures measured on these two scales.
Solution:
Since both scales are absolute, their zero points correspond to absolute zero. This means that a certain temperature difference on one scale will correspond to the same temperature difference on the other scale.
Let’s denote:
* TA: Temperature on scale A
* TB: Temperature on scale B
We know that the difference between the triple point of water and absolute zero is the same on both scales. So, the proportion as
(TA / 200 A) = (TB / 350 B)
Cross-multiplying:
350 TA = 200 TB
Hnce, the relation between TA and TB is:
TA = (4/7) TB
3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Ans : Understanding the Problem:
We are given the resistance of a resistor at two different temperatures.
We need to find the temperature at which the resistance is 123.4 ohms.
Using the Temperature Coefficient of Resistance:
Given,
relationship between (R) and (T)resistance and temperature respectively is :
R = R₀[1 + α(T – T₀)]
Where:
R₀= resistance at a reference temperature T₀.
α = temperature coefficient of resistance.
Calculating the Temperature Coefficient (α):
We are given two cases:
Case (i): R₁ = 165.5 Ω, T₁ = 600.5 K
Case (ii): R₂ = 123.4 Ω, T₂ = ?
Using the formula for case (i):
165.5 Ω = 101.6 Ω [1 + α(600.5 K – 273.16 K)]
Solving for α:
α = (165.5 – 101.6) / (101.6 × 327.34) = 63.9 / 32734
Finding the Temperature (T₂):
Using the formula for case (ii):
123.4 Ω = 101.6 Ω [1 + α(T₂ – 273.16)]
Substituting the value of α:
123.4 Ω = 101.6 Ω [1 + (63.9/32734)(T₂ – 273.16)]
Solving for T₂:
T₂ = [(123.4 – 101.6) × 327.34 / 63.9] + 273.16
= 384.83 K
Therefore, the temperature at which the resistance is 123.4 ohms is 384.83 K.
4 .Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15 Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?
Ans : (a) The triple point of water is a standard fixed point because it represents a unique state where water can exist simultaneously as a solid, liquid, and gas. This point is highly reproducible and does not depend on external pressure, unlike the melting and boiling points of water, which can vary with pressure.
Using the melting point of ice and the boiling point of water as fixed points has the disadvantage of being dependent on atmospheric pressure. These points can vary slightly at different atmospheric pressures, leading to inaccuracies in temperature measurements.
(b) The other fixed point on the Kelvin scale is absolute zero, which is the theoretical temperature at which all molecular motion ceases. It is given a value of absolute zero, or 0 Kelvin.
(c) The relation between the Kelvin and Celsius scales is given by:
tc = T – 273.15
The reason for subtracting 273.15 is to account for the difference in the zero points of the two scales. The Kelvin scale starts at absolute zero, while the Celsius scale starts at the freezing point of water, which is 273.15 K.
(d) To find the temperature of the triple point of water on this new absolute scale, we need to consider the ratio of the temperature intervals.
180 degrees Fahrenheit is equal to 100 degrees Celsius.. So, 1 degree Celsius is equal to 1.8 degrees Fahrenheit.
The triple point of water on the Kelvin scale is 273.16 K. To convert this to the new absolute scale, we multiply by the ratio of the temperature intervals:
Temperature on new scale = 273.16 K * (1.8 °F/°C)
= 491.688 units.
Therefore, the temperature of the triple point of water on this new absolute scale is approximately 491.69 units.
5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
Temperature Pressure thermometer A Pressure thermometer B
Triple-point of water 1.250 × 105 Pa 0.200 × 105 Pa
Normal melting point 1.797 × 105 Pa 0.287 × 105 Pa of sulphur
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
Ans : Analyzing the Ideal Gas Thermometers
(a) Absolute Temperature of Normal Melting Point of Sulphur
Using Thermometer A:
At the triple point of water (273.16 K), the pressure is 1.250 x 10^5 Pa.
At the normal melting point of sulfur, the pressure is 1.797 x 10^5 Pa.
Assuming ideal gas behavior, we can use the relation:
P/T = constant
Therefore, for thermometer A:
(1.250 x 10^5 Pa) / 273.16 K = (1.797 x 10^5 Pa) / T_A
Solving for T_A, we get:
T_A ≈ 392.5 K
Using Thermometer B:
At the triple point of water (273.16 K), the pressure is 0.200 x 10^5 Pa.
At the normal melting point of sulfur, the pressure is 0.287 x 10^5 Pa.
Similarly, for thermometer B:
(0.200 x 10^5 Pa) / 273.16 K = (0.287 x 10^5 Pa) / T_B
Solving for T_B, we get:
T_B ≈ 389.0 K
(b) Reason for Discrepancy and Improvement
The slight difference in the absolute temperature readings from the two thermometers can be attributed to the non-ideal behavior of the gases. Real gases deviate from ideal gas behavior, especially at higher pressures and lower temperatures.
To reduce the discrepancy, we can:
Use a gas that behaves more ideally: Gases like hydrogen and helium, which have smaller molecular sizes and weaker intermolecular forces, tend to exhibit more ideal behavior.
Extrapolate to zero pressure: By measuring the pressure at different temperatures and extrapolating the pressure-temperature graph to zero pressure, we can obtain a more accurate temperature reading, as ideal gas behavior is more closely approached at lower pressures.
Use multiple gas thermometers: By using multiple gas thermometers with different gases and comparing their readings, we can get a more accurate and reliable measurement of temperature.
By implementing these strategies, we can minimize the effects of non-ideal gas behavior and obtain more accurate temperature measurements.
6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1 .
Ans : *Understanding the Problem:**
We have a steel tape calibrated at 27°C. We’re measuring a steel rod at a higher temperature, leading to expansion of both the tape and the rod. We need to find the actual length of the rod on both hot and cold days.
**Solution:**
1. Expansion of the Steel Tape:
The increase in length of the tape (ΔL₁) is given by:
ΔL₁ = αL₀ΔT
Where:
* α is the coefficient of linear expansion of steel (1.20 × 10⁻⁵ K⁻¹)
* L₀ is the original length of the tape (1 m)
* ΔT is the change in temperature (45°C – 27°C = 18°C)
Calculating ΔL₁:
ΔL₁ = (1.20 × 10⁻⁵ K⁻¹)(1 m)(18 K) = 2.16 × 10⁻⁴ m
So, the actual length of the tape at 45°C is:
L₁ = L₀ + ΔL₁ = 1 m + 2.16 × 10⁻⁴ m = 1.000216 m
2. Actual Length of the Steel Rod at 45°C:
The measured length of the rod (L₂) is 63.0 cm = 0.63 m. This is the length of the rod relative to the expanded tape.
To find the actual length of the rod (L₃) at 45°C, we need to account for the expansion of the tape:
L₃ = L₂ / L₁ = 0.63 m / 1.000216 ≈ 0.6299 m
**3. Length of the Rod at 27°C:**
Since both the rod and the tape are at the same temperature (27°C), there is no difference in their expansion. So, the measured length of the rod (0.63 m) is the actual length at 27°C.
Therefore, the actual length of the steel rod on the hot day is approximately 0.6299 m, and on the cold day, it is 0.63 m..
7 .A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : αsteel = 1.20 × 10–5 K–1 .
Ans : Understanding the Problem:
We have a steel shaft with an initial diameter of 8.70 cm at a temperature of 27°C.
The diameter of the shaft decreases to 8.69 cm due to a change in temperature.
We need to find the final temperature of the shaft.
Calculations:
1. Change in Diameter:
Determine change in diameter (ΔD)
i.e. ΔD = D₁α(T₂ – T₁)
Where:
D₁ is the initial diameter (8.70 cm)
α is the coefficient of linear expansion for steel (1.20 x 10^-5 °C^-1)
T₂ is the final temperature
T₁ is the initial temperature (27°C)
2. Finding the Final Temperature (T₂):
Substituting the values into the formula:
8.69 cm – 8.70 cm = 8.70 cm × 1.20 × 10^-5 °C^-1 × (T₂ – 27°C)
Simplifying and solving for T₂:
-0.01 cm = 1.044 × 10^-4 °C^-1 × (T₂ – 27°C)
T₂ – 27°C = -0.01 cm / (1.044 × 10^-4 °C^-1) ≈ -95.8 °C
T₂ ≈ 27°C – 95.8°C ≈ -68.8°C
≈ -69°C
Therefore, the final temperature of the steel shaft is approximately -69°C.
8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1 .
Ans : *Understanding the Problem:**
We have a copper sheet with a hole in it. As the sheet is heated, it expands, which will also cause the hole to expand.
Determine change in the diameter of the hole.
Solution:
1. Calculate the Change in Diameter:
The change in the diameter of the hole (ΔD) is proportional to the original diameter (D₀) and the change in temperature (ΔT):
ΔD = αD₀ΔT
Where:
* α is the coefficient of linear expansion of copper (1.70 × 10⁻⁵ K⁻¹)
* D₀ is the original diameter of the hole (4.24 cm = 0.0424 m)
* ΔT is the change in temperature (227°C – 27°C = 200°C)
Substituting the values:
ΔD = (1.70 × 10⁻⁵ K⁻¹)(0.0424 m)(200 K) ≈ 1.44 × 10⁻⁴ m
2. Calculate the Final Diameter:
The final diameter (D) is the sum of the original diameter and the change in diameter:
D = D₀ + ΔD = 0.0424 m + 1.44 × 10⁻⁴ m ≈ 0.042544 m
Converting to centimeters:
D ≈ 4.2544 cm
Therefore, the change in the diameter of the hole is approximately 0.0144 cm, and the final diameter is approximately 4.2544 cm.
9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
Ans : Understanding the Problem:
We have a steel rod with a length (L) of 1.8 meters.
The temperature of the rod changes from 27°C to -39°C (a change of -66°C).
We need to calculate the force exerted on the rod due to this temperature change.
Calculations:
1. Change in Length (ΔL):
Given,
ΔL = αLΔT
Where:
α is the coefficient of linear expansion of steel (2.0 x 10^-5 K^-1)
ΔT is the change in temperature (-66°C)
2. Force Exerted (F):
The force exerted on the rod is related to the change in length by Young’s modulus (Y):
Y =
(F × L)
————
(A × ΔL)
Where:
A = cross-sectional area of the rod
F = force exerted
Rearranging the formula to solve for F:
F = (Y × A × ΔL)
—————–
L
Substituting the values:
F = (0.91 × 10^11 Pa) × (π/4) × (2 × 10^-3 m)^2 × (2.0 × 10^-5 K^-1) × (-66°C)
Calculating the force:
F ≈ -3.77 × 10^2 N
Therefore, the force exerted on the steel rod is approximately -377 N.
10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1 ).
Ans : Understanding the Problem:
We have a composite rod made of brass and steel, each 50 cm long and 3 mm in diameter.
The temperature of the rod increases from 40°C to 250°C.
We need to find the total change in length of the composite rod.
Calculations:
1. Change in Length of Brass:
The change in length of the brass section (ΔL_brass) is calculated using the formula:
ΔL_brass = α_brass × L_brass × ΔT
Where:
α_brass is the coefficient of linear expansion of brass (2 x 10^-5 °C^-1)
L_brass is the length of the brass section (50 cm)
ΔT is the change in temperature (210°C)
Substituting the values:
ΔL_brass = 2 x 10^-5 °C^-1 × 50 cm × 210°C
= 0.21 cm
2. Change in Length of Steel:
Similarly, the change in length of the steel section (ΔL_steel) is calculated using:
ΔL_steel = α_steel × L_steel × ΔT
Where:
α_steel is the coefficient of linear expansion of steel (1.2 x 10^-5 °C^-1)
L_steel is the length of the steel section (50 cm)
Substituting the values:
ΔL_steel = 1.2 x 10^-5 °C^-1 × 50 cm × 210°C = 0.126 cm
≈ 0.13 cm
3. Total Change in Length:
The total change in length (ΔL) is the sum of the changes in the brass and steel sections:
ΔL = ΔL_brass + ΔL_steel = 0.21 cm + 0.13 cm
= 0.34 cm
Conclusion:
The total change in length of the composite rod is 0.34 cm. Since the rod is not clamped, no thermal stress develops at the junction between the brass and steel sections.
11. The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature ?
Ans : Understanding the Problem:
We are given a material with a coefficient of volume expansion (γ) of 49 x 10^-5 °C^-1.
The temperature of the material increases by 30°C.
We need to find the fractional change in the density of the material.
Calculations:
1. Change in Volume:
The change in volume (ΔV) due to the temperature change is given by:
ΔV = γVΔT
Where:
V = initial volume of the material
ΔT = change in temperature (30°C)
The final volume (V’) can be expressed as:
V’ = V + ΔV = V + γVΔT = V(1 + γΔT)
2. Change in Density:
(ρ) is defined as (m) per unit (V):
ρ = m/V
The initial density (ρ) is:
ρ = m/V
The final density (ρ’) is:
ρ’ = m/V’ = m / [V(1 + γΔT)] = ρ / (1 + γΔT)
3. Fractional Change in Density:
Given,
fractional change in density
Fractional change = (ρ – ρ’) / ρ
Substituting the expression for ρ’:
Fractional change = (ρ – ρ / (1 + γΔT)) / ρ
Simplifying:
Fractional change = 1 – 1 / (1 + γΔT)
Calculating the numerical value:
Fractional change = 1 – 1 / (1 + 49 x 10^-5 x 30)
≈ 0.0145
Therefore, the fractional change in density of the material is approximately 0.0145.
12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1 .
Ans : *Understanding the Problem:**
We have a drilling machine that is used to drill an aluminum block. We’re given the power of the machine, the mass of the block, and its specific heat capacity. We need to find the rise in temperature of the block, assuming 50% of the power is used to heat the block.
1. Determine Energy Supplied to the Block:
* Power of the machine = 10 kW = 10,000 W
* Time
= 2.5 minutes
= 150 seconds
* 50% of the power is used to heat the block.
* So, power used to heat the block = 50% of 10,000 W = 5000 W
* Energy supplied to the block (Q) = Power × Time = 5000 W × 150 s = 750,000 J
2. Calculate the Temperature Rise:
The heat energy (Q) supplied to the block is used to raise its temperature:
Q = mcΔT
Where:
* m is the mass of the block (8 kg = 8000 g)
* c is the specific heat capacity of aluminum (0.91 J/g K)
* ΔT is the change in temperature
Solving for ΔT:
ΔT = Q / (mc) = 750,000 J / (8000 g × 0.91 J/g K)
≈ 103 K
Therefore, the rise in temperature of the aluminum block is approximately 103 K.
13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1 ).
Ans : Understanding the Problem:**
We have a hot copper block that is placed on ice. The heat released by the cooling copper block will be used to melt the ice. We need to calculate the maximum mass of ice that can be melted.
Solution:
1. Calculate the Heat Released by the Copper Block:
The heat released by the copper block (Q) can be calculated using the formula:
Q = mcΔT
Where:
* m is the mass of the copper block (2.5 kg = 2500 g)
* c is the specific heat capacity of copper (0.39 J/g K)
* ΔT is the change in temperature (500°C – 0°C = 500 K)
Substituting the values:
Q = 2500 g × 0.39 J/g K × 500 K = 487,500 J
2. Calculate the Mass of Ice Melted:
The heat released by the copper block is used to melt the ice. The mass of ice melted (m_ice) can be calculated using the latent heat of fusion (L):
Q = m_ice × L
Where:
* L is the latent heat of fusion of water (335 J/g)
Solving for m_ice:
m_ice = Q / L = 487,500 J / 335 J/g
≈ 1455 g
Therefore, the maximum amount of ice that can be melted is approximately 1.455 kg.
14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
Ans : Understanding the Problem:
A metal block with a mass of 200 g is heated to 150°C and then placed in a calorimeter containing 150 g of water at 27°C.
The final temperature of the water and calorimeter is 40°C.
Determine specific heat capacity of the metal.
Calculations:
1. Heat Lost by the Metal Block:
The heat lost by the metal block (Q_metal) is given by:
Q_metal = m_metal × C_metal × ΔT_metal
Where:
m_metal is the mass of the metal block (200 g)
C_metal is the specific heat capacity of the metal (unknown)
ΔT_metal is the change in temperature of the metal (150°C – 40°C = 110°C)
2. Heat Gained by the Water:
The amount of heat absorbed by the water (Q_water) is given by:
Q_water = m_water × C_water × ΔT_water
Where:
The mass of the water, m_water, is 150 grams.
C_water is the specific heat capacity of water (1 cal/g°C)
ΔT_water is the change in temperature of the water (40°C – 27°C = 13°C)
3. Heat Gained by the Calorimeter:
The heat gained by the calorimeter (Q_calorimeter) is given by:
Q_calorimeter = W × ΔT_calorimeter
Where:
W is the water equivalent of the calorimeter (25 g)
ΔT_calorimeter is the change in temperature of the calorimeter (13°C)
4. Equating Heat Loss and Heat Gain:
According to the principle of calorimetry, the heat lost by the metal block is equal to the heat gained by the water and the calorimeter:
Q_metal = Q_water + Q_calorimeter
Substituting the values:
200 g × C_metal × 110°C = (150 g × 1 cal/g°C × 13°C) + (25 g × 13°C)
5. Calculating the Specific Heat Capacity of the Metal:
Solving for C_metal:
C_metal = (175 × 13) / (200 × 110) = 0.1 cal/g°C
≈ 0.43 J/gK
Therefore, the specific heat capacity of the metal is 0.1 cal/g°C or 0.43 J/gK.
15. Given below are observations on molar specific heats at room temperature of some common gases
Gas Molar specific heat (Cv ) (cal mo1–1 K–1)
Hydrogen 4.87
Nitrogen 4.97
Oxygen 5.02
Nitric oxide 4.99
Carbon monoxide 5.01
Chlorine 6.17
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
Ans : *The gases listed in the table are diatomic, not monoatomic. Diatomic gases, like hydrogen, nitrogen, oxygen, and chlorine, have a molar specific heat of approximately 5/2 R, which is around 4.95 cal/mol/K. This value agrees well with the table’s data, except for chlorine.
Monoatomic gases, like helium and neon, only have translational motion. Diatomic gases, on the other hand, can also undergo rotational and vibrational motion. To raise the temperature of a diatomic gas by 1°C, energy is required to increase not only its translational energy but also its rotational and vibrational energy. This is why diatomic gases have a higher molar specific heat than monoatomic gases.
Chlorine has a slightly higher molar specific heat compared to other diatomic gases at room temperature. This suggests that, in addition to translational and rotational motion, chlorine molecules also undergo vibrational motion at room temperature. Other diatomic gases, like hydrogen, nitrogen, and oxygen, typically only exhibit rotational motion at room temperature. This difference in vibrational behavior explains the higher molar specific heat of chlorine.
16. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1 .
Ans : Understanding the Problem:
We are given a mass of water (30 kg) that cools from 101°F to 98°F.
We need to calculate the rate of evaporation of water at this temperature.
Calculations:
Temperature Change:
The decrease in temperature (ΔT) is:
ΔT = 101°F – 98°F = 3°F = 3 × (5/9) °C
= 1.67°C
Heat Lost:
The heat lost by the water (Q) is calculated using the formula:
Q =
mass* specific heat capacity *ΔT
Specific heat capacity of water = 1000 cal/kg°C
Q = 30 kg × 1000 cal/kg°C × 1.67°C = 50100 cal
Mass of Water Evaporated:
The heat lost by the water is used to evaporate some of the water.
The amount of water evaporated (m’) can be calculated using the formula:
m’ = Q / L
Latent heat of vaporization (L) = 580 × 10^3 cal/kg
m’ = 50100 cal / (580 × 10^3 cal/kg)
= 0.086 kg
Rate of Evaporation:
This evaporation process takes 20 minutes.
So, the rate of evaporation is:
Rate of evaporation = 0.086 kg / 20 min = 86 g / 20 min
= 4.3 g/min
Therefore, the rate of evaporation of water is 4.3 grams per minute.
17. A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1.
[Heat of fusion of water = 335 × 103 J kg–1]
Ans : *Understanding the Problem:**
We’re asked to calculate the amount of ice remaining in a thermacole icebox after a certain time, given the outside temperature, the thickness of the box, and the thermal conductivity of the material.
1. Calculate the Surface Area:
The surface area of the cube-shaped icebox is:
A = 6 × side² = 6 × (0.3 m)²
= 0.54 m²
2. Calculate the Rate of Heat Flow:
The rate of heat flow (Q/t) through the walls of the icebox can be calculated using the formula:
Q/t = (kAΔT) / d
Where:
* k is the thermal conductivity of the thermacole (0.01 J/s m K)
* A is the surface area (0.54 m²)
* ΔT is the temperature difference between the inside and outside of the box (45°C – 0°C = 45 K)
* d is the thickness of the wall (0.05 m)
Substituting the values:
Q/t = (0.01 J/s m K × 0.54 m² × 45 K) / 0.05 m
= 4.86 J/s
3. Calculate the Heat Required to Melt the Ice:
The heat required to melt 4 kg of ice is:
Q_melt = mass × latent heat of fusion = 4 kg × 335 × 10³ J/kg
= 1.34 × 10⁶ J
4. Determine Time Taken to Melt the Ice:The time taken to melt the ice can be calculated by dividing the heat required to melt the ice by the rate of heat flow:
Time = Q_melt / (Q/t) = (1.34 × 10⁶ J) / (4.86 J/s) ≈ 275,216 seconds
≈ 76.45 hours
5. Determine the Amount of Ice Remaining:
Since 6 hours is less than the time calculated to melt all the ice, some ice will remain.
To find the amount of ice melted in 6 hours:
Heat absorbed in 6 hours = (4.86 J/s) × (6 × 3600 s) = 104,976 J
Mass of ice melted = Heat absorbed / Latent heat of fusion = 104,976 J / (335 × 10³ J/kg) ≈ 0.313 kg
Therefore, the amount of ice remaining after 6 hours is:
4 kg – 0.313 kg
≈ 3.687 kg
18. A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 10^9 J s–1 m–1 K–1 ; Heat of vaporisation of water = 2256 × 10^3 J kg–1 .
Ans : Understanding the Problem:
We are given a boiler with a base made of a material with thermal conductivity
K = 109 J/(s m°C).
The base has an area A = 0.15 m² and thickness d = 1 cm = 0.01 m.
Water is boiling at a rate of 6.0 kg/min.
Find the temperature of the boiler’s base in contact with the stove.
Assumptions:
We assume steady-state heat conduction through the base of the boiler.
We assume that all the heat conducted through the base is used to vaporize the water.
Calculations:
Heat Flow Through the Base:
The rate of heat flow (Q) through the base is given by Fourier’s law of heat conduction:
Q = (KA(T1 – T2)) / d
Where:
T1 = temperature of the part of the boiler in contact with the stove.
T2 = temperature of the water (100°C).
Substituting the given values:
Q = (109 J/(s m°C) × 0.15 m² × (T1 – 100°C)) / 0.01 m
Simplifying:
Q = 1635(T1 – 100) J/s
Heat Required for Vaporization:
The rate of heat required to vaporize the water (Q) is given by:
Q = M × L
Where:
M denote mass flow rate of water (0.1 kg/s)
L is the latent heat of vaporization of water (2256 × 10^3 J/kg)
Substituting the values:
Q = 0.1 kg/s × 2256 × 10^3 J/kg = 225600 J/s
Equating Heat Flow and Heat Required:
Since all the heat conducted through the base is used to vaporize the water, we can equate the two expressions for Q:
1635(T1 – 100) = 225600
Solving for T1:
T1 – 100 = 225600 / 1635 ≈ 138
Therefore, T1 = 138 + 100
= 238°C.
Conclusion:
The part of the boiler in contact with the stove is at a temperature of 238°C.
19. Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
Ans : (a) Reflectivity and Emissivity:A body with a large reflectivity reflects most of the incident radiation. This means it absorbs less radiation, and hence, it emits less radiation as well. A body that reflects most of the incident radiation is a poor emitter of radiation.
(b) Thermal Conductivity:
Brass is a good conductor of heat, while wood is a poor conductor. When you touch a cold object, heat is transferred from your hand to the object. Brass, being a good conductor, quickly conducts heat away from your hand, making it feel colder. Wood, being a poor conductor, conducts heat slowly, making it feel less cold.
(c) Black Body Radiation:
An ideal black body absorbs and emits radiation perfectly. A red-hot iron piece in the open is not a perfect black body. It reflects some of the incident radiation, leading to a lower apparent temperature when measured with an optical pyrometer calibrated for a black body. Inside a furnace, the iron piece is surrounded by hot walls, which radiate heat onto it, making it behave more like a black body.
(d) Earth’s Atmosphere and Greenhouse Effect:
The Earth’s atmosphere acts like a blanket, trapping heat and preventing it from escaping into space. . This greenhouse effect keeps the Earth’s temperature warm and habitable. Without the atmosphere, the Earth would lose heat rapidly through radiation, leading to extremely cold temperatures.
(e)Steam Heating Systems:
Steam heating systems are more efficient than hot water systems because of the latent heat of vaporization. When steam condenses, it releases a large amount of latent heat, which is transferred to the surroundings. Hot water, on the other hand, only transfers heat through convection.
20. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Ans : Understanding the Problem:
We are given a body cooling according to Newton’s law of cooling.
The body cools from 80°C to 50°C in 5 minutes when the surrounding temperature is 20°C.
We need to find the time taken for the body to cool from 60°C to 30°C under the same conditions.
Applying Newton’s Law of Cooling:
Newton’s law of cooling states that the rate of cooling of an object is directly proportional to the temperature difference between the object and its surroundings.
Calculations:
Initial Cooling:
Average temperature = (80°C + 50°C) / 2 = 65°C
Temperature difference (ΔT1) = 65°C – 20°C = 45°C
Cooling time (t1) = 5 minutes
Change in temperature (ΔT) = 30°C
Using Newton’s law of cooling:
ΔT / t1 = K * ΔT1
where K is a constant of proportionality.
Substituting the values:
30°C / 5 min = K * 45°C
(Equation 1)
Second Cooling:
Average temperature = (60°C + 30°C) / 2 = 45°C
Temperature difference (ΔT2) = 45°C – 20°C = 25°C
Cooling time (t2) = ?
Change in temperature (ΔT) = 30°C
Using Newton’s law of cooling again:
ΔT / t2 = K * ΔT2
Substituting the values:
30°C / t2 = K * 25°C
(Equation 2)
Comparing the Equations:
Dividing Equation 1 by Equation 2:
(30°C / 5 min) / (30°C / t2) = (K * 45°C) / (K * 25°C)
Simplifying:
t2 / 5 = 9 / 5
Solving for t2:
t2 = 9 minutes
Therefore, the body will take 9 minutes to cool from 60°C to 30°C under the given conditions.
