Question:

A brass wire 1.8 m long at 270C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of −39oC, what is the tension developed in the wire, if its diameter is 2.0mm? Co-efficient of linear expansion of brass =2.0×10^−5K^−1; Young’s modulus of brass =0.91×10^11 Pa

Answer:

Step 1: Calculate the change in length of the wire due to cooling. Change in temperature = 270 - (-39) = 309 Change in length = (2.0 x 10^-5 x 309) x 1.8 = 0.056 m

Step 2: Calculate the change in tension in the wire due to cooling. Change in tension = (0.91 x 10^11 x 0.056) / (2 x 10^-3) = 5.064 x 10^7 Pa

Question:

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500oC and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper=0.39Jg^−1K^−1; heat of fusion of water =335Jg^−1)

Answer:

1. Calculate the amount of heat energy required to raise the temperature of the copper block from 0oC to 500oC.

Q = (2.5 kg)(0.39 J/gK)(500oC - 0oC) = 487.5 kJ

2. Calculate the amount of heat energy required to melt the ice.

Q = (335 J/g)(mass of ice)

3. Set the two equations equal to each other and solve for the mass of ice.

487.5 kJ = (335 J/g)(mass of ice)

mass of ice = 1.45 kg

Question:

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

Answer:

(a) A body with large reflectivity is a poor emitter because it reflects most of the light and heat energy that it receives, rather than emitting it.

(b) A brass tumbler feels much colder than a wooden tray on a chilly day because brass is a better conductor of heat than wood, so it absorbs heat from the environment more quickly.

(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 because the iron piece in the open is not an ideal black body, so its radiation is not the same as an ideal black body radiation. When the iron piece is in the furnace, it is surrounded by other objects that are radiating at the same temperature, so its radiation is more like that of an ideal black body.

(d) The earth without its atmosphere would be inhospitably cold because the atmosphere acts as a blanket, trapping heat from the sun and preventing it from escaping into outer space.

(e) Heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water because steam is a better conductor of heat than water, so it can transfer heat more quickly.

Question:

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 273.16K, 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 ?

Answer:

Step 1: Identify the given information: Ro = 101.6 Ω, To = 273.16K, R = 165.5 Ω, T = 600.5 K, and R = 123.4 Ω

Step 2: Rearrange the equation to solve for T: R = Ro[1 + α(T − To)]

Step 3: Substitute the known values: 123.4 = 101.6[1 + α(T − 273.16)]

Step 4: Solve for T: T = (123.4 - 101.6)/α + 273.16

Step 5: Substitute the known values for R and T: T = (123.4 - 101.6)/(165.5 - 101.6)/(600.5 - 273.16) + 273.16

Step 6: Calculate the value of T: T = 393.2 K

Question:

A large steel wheel is to be fitted on to a shaft of the same material. At 27oC the outer diameter of the shaft is 8.70cm and the diameter of the central hole in the wheel is 8.69cm. 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 the constant over the required temperature range : αsteel​=1.20×10^−5K^−1

Answer:

1. Calculate the change in diameter of the shaft due to cooling using the coefficient of linear expansion: Change in diameter = αsteel × (change in temperature) × (original diameter) Change in diameter = 1.20 × 10-5 K-1 × (T2 - 27) × 8.70 cm

2. Calculate the new diameter of the shaft after cooling to the required temperature: New diameter = Original diameter + Change in diameter New diameter = 8.70 cm + (1.20 × 10-5 K-1 × (T2 - 27) × 8.70 cm)

3. Calculate the temperature required for the wheel to slip on the shaft: New diameter = 8.69 cm 8.69 cm = 8.70 cm + (1.20 × 10-5 K-1 × (T2 - 27) × 8.70 cm) T2 = 27 + [(8.69 cm - 8.70 cm)/(1.20 × 10-5 K-1 × 8.70 cm)] T2 = 27 + [(-0.01 cm)/(1.20 × 10-5 K-1 × 8.70 cm)] T2 = 27 - (8.33 × 10-4 K) T2 = 26.99167 K

Therefore, the required temperature for the wheel to slip on the shaft is 26.99167 K.

Question:

A child running a temperature of 1010F is given an antipyrin (i.e. a medicine that lowers fever ) which causes an increases in the rate of evaporation of sweat from his body. If the fever is brought down to 98oF in 20 min, 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 580calg^−1.

Answer:

1. Calculate the total amount of heat lost from the body of the child due to the extra evaporation caused by the drug:

Heat Lost = (1010F - 98F) x 30 kg x specific heat of water x 20 min

2. Calculate the average rate of extra evaporation caused by the drug:

Average Rate of Extra Evaporation = Heat Lost / (20 min x latent heat of evaporation of water at that temperature)

3. The average rate of extra evaporation caused by the drug is:

Average Rate of Extra Evaporation = (1010F - 98F) x 30 kg x specific heat of water x 20 min / (20 min x 580calg^−1)

Question:

A brass boiler has a base area of 0.15m^2 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 =109Js^−1m^−1K^−1; Heat of vaporisation of water =2256×10^3Jkg^−1.

Answer:

Step 1: Calculate the rate of heat transfer from the flame to the boiler.

The rate of heat transfer from the flame to the boiler is given by:

Q = (Thermal conductivity of brass) x (Area of base) x (Thickness of boiler) x (Temperature difference)

Q = (109 Js^−1m^−1K^−1) x (0.15 m^2) x (1.0 cm) x (T2 - T1)

Step 2: Calculate the rate of heat needed to boil the water.

The rate of heat needed to boil the water is given by:

Q = (Heat of vaporisation of water) x (Mass of water boiled)

Q = (2256×10^3 Jkg^−1) x (6.0 kg/min)

Step 3: Set the two equations equal to each other and solve for the temperature difference.

(109 Js^−1m^−1K^−1) x (0.15 m^2) x (1.0 cm) x (T2 - T1) = (2256×10^3 Jkg^−1) x (6.0 kg/min)

T2 - T1 = (2256×10^3 Jkg^−1) x (6.0 kg/min) / [(109 Js^−1m^−1K^−1) x (0.15 m^2) x (1.0 cm)]

T2 - T1 = 17,835 K

Step 4: Calculate the temperature of the part of the flame in contact with the boiler.

The temperature of the part of the flame in contact with the boiler is given by:

T2 = T1 + (T2 - T1)

T2 = T1 + 17,835 K

Therefore, the temperature of the part of the flame in contact with the boiler is T1 + 17,835 K.

Question:

The coefficient of volume expansion of glycerine is 49×10^−5K^−1. What is the fractional change in its density for a 30oC rise in temperature? A 1.5×10^−2 B 2×10^−4 C 3.5×10^−3 D 2.5×10^−2

Answer:

Step 1: The coefficient of volume expansion of glycerine is 49×10^−5K^−1.

Step 2: Calculate the fractional change in its density for a 30oC rise in temperature.

Step 3: The fractional change in density = (49×10^−5K^−1)(30)

Step 4: The fractional change in density = 1.47×10^−3

Step 5: Compare the answer with the given options and select the correct option.

Answer: C 3.5×10^−3

Question:

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg.Find the rise in temperature of the block in 2.5 minutes, assuming 50% power is used up in heating the machine itself or lost to the surroundings. (Specific heat of aluminium =0.90Jg^−10−1) A 100 B 103 C 150 D 155

Answer:

Answer: D 155

Step 1: Calculate the power used to heat the block:

Power used to heat the block = 10 kW - (50% of 10 kW) = 5 kW

Step 2: Calculate the amount of heat energy generated:

Heat energy generated = 5 kW x 2.5 minutes = 12.5 kJ

Step 3: Calculate the amount of heat absorbed by the block:

Amount of heat absorbed by the block = 12.5 kJ

Step 4: Calculate the mass of the block:

Mass of the block = 8 kg

Step 5: Calculate the specific heat of aluminium:

Specific heat of aluminium = 0.90 Jg^−10−1

Step 6: Calculate the rise in temperature of the block:

Rise in temperature of the block = (12.5 kJ) / (8 kg x 0.90 Jg^−10−1) = 155°C

Question:

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 0oC and 100oC 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 relate 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 ?

Answer:

A. (a) The triple-point of water is a standard fixed point in modern thermometry because it is the temperature at which all three states of water (solid, liquid, and gas) can coexist in equilibrium. Taking the melting point of ice and the boiling point of water as standard fixed points is inaccurate because these points vary depending on the atmospheric pressure, which can change significantly from one location to another.

B. (b) The other fixed point on the Kelvin scale is absolute zero, which is assigned the number 0 K.

C. (c) We have 273.15 in this relation because it is the temperature of the triple-point of water on the Celsius scale.

D. (d) The temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale is 32°F.

Question:

The coefficient of volume expansion of glycerine is 49×10^−5K^−1. What is the fractional change in its density for a 30oC rise in temperature? A 1.5×10^−2 B 2×10^−4 C 3.5×10^−3 D 2.5×10^−2

Answer:

Answer: A 1.5×10^−2

Step 1: Understand the question.

The question is asking for the fractional change in the density of glycerine for a 30oC rise in temperature.

Step 2: Calculate the fractional change in the density.

The coefficient of volume expansion of glycerine is 49×10^−5K^−1. This means that for every 1K rise in temperature, the volume of glycerine will increase by 49×10^−5.

For a 30oC rise in temperature, the volume of glycerine will increase by (30 × 49×10^−5) = 1.47×10^−3.

The fractional change in the density of glycerine is (1.47×10^−3/1) = 1.47×10^−3.

Step 3: Select the correct answer.

The correct answer is A 1.5×10^−2.

Question:

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg.Find the rise in temperature of the block in 2.5 minutes, assuming 50% power is used up in heating the machine itself or lost to the surroundings. (Specific heat of aluminium =0.90Jg^−10−1) A 100 B 103 C 150 D 155

Answer:

Answer: D 155

Step 1: Calculate the power used to heat the aluminium block

Power used to heat the aluminium block = 10 kW - (50% x 10 kW) = 5 kW

Step 2: Calculate the amount of heat energy required to raise the temperature of the aluminium block

Amount of heat energy = (mass of aluminium block x specific heat of aluminium x change in temperature)

= (8 kg x 0.90 Jg^-1K^-1 x ΔT)

Step 3: Calculate the change in temperature

Change in temperature (ΔT) = (Amount of heat energy / (mass of aluminium block x specific heat of aluminium))

= (5 kW x 2.5 minutes x 60 seconds/minute) / (8 kg x 0.90 Jg^-1K^-1)

= 155 K

Question:

A ’thermacole’ icebox is a cheap and 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 45oC, and co-efficient of thermal conductivity of thermacole is 0.01Js^−1m^−1K^−1. [Heat of fusion of water =335×10^3Jkg^−1]

Answer:

1. Calculate the surface area of the icebox: A = 6*(30 cm)^2 = 5400 cm^2

2. Calculate the rate of heat transfer through the icebox: Q = kA(T1 - T2)/d = 0.01 Js^−1m^−1K^−1 * 5400 cm^2 * (45oC - 0oC)/5.0 cm = 2592 Js^−1

3. Calculate the amount of heat lost by the ice: Q = m*L = 4.0 kg * 335×10^3Jkg^−1 = 1340×10^3J

4. Calculate the amount of ice remaining after 6 h: Q = mL 1340×10^3J = m335×10^3Jkg^−1 m = 4.0 kg - (1340×10^3J/(335×10^3Jkg^−1)) * (6 h/3600 s) = 3.7 kg

Question:

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​ ?

Answer:

1. Determine the difference between the two triple points. Answer: The difference between the two triple points is 150.

2. Calculate the ratio between the two triple points. Answer: The ratio between the two triple points is 200/350 = 0.571.

3. Calculate the relation between TA and TB. Answer: The relation between TA and TB is TA = 0.571 x TB.

Question:

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 250oC, if the original lengths are at 40.0oC? 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^−5K^−1, steel =1.2×10^−5K^−1)

Answer:

1. Calculate the change in length of the brass rod at 250oC using the formula ΔL = L0 * α * (T2 - T1), where L0 is the initial length of the rod, α is the coefficient of linear expansion, and T2 and T1 are the final and initial temperatures, respectively.

ΔL = 50 cm * 2.0 x 10^-5 K^-1 * (250 - 40) = 0.064 cm

2. Calculate the change in length of the steel rod at 250oC using the same formula.

ΔL = 50 cm * 1.2 x 10^-5 K^-1 * (250 - 40) = 0.038 cm

3. Calculate the total change in length of the combined rod at 250oC by adding the change in length of the brass rod and the steel rod.

ΔL = 0.064 cm + 0.038 cm = 0.102 cm

4. Determine if there is thermal stress developed at the junction by calculating the difference in the change in length of the brass rod and the steel rod.

ΔL = 0.064 cm - 0.038 cm = 0.026 cm

Yes, there is thermal stress developed at the junction due to the difference in the change in length of the brass rod and the steel rod.

Question:

In an experiment on the specific heat of a metal a 0.20 kg block of the metal at 150oC is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150cm^3 of water at 27oC. The final temperature is 40oC. 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 ?

Answer:

1. Calculate the heat gained by the water: Q = mcΔT = (0.025 kg)(4.186 J/g°C)(40oC - 27oC) = 2.7 J

2. Calculate the heat lost by the metal: Q = mcΔT = (0.20 kg)(4.186 J/g°C)(150oC - 40oC) = 462.4 J

3. Calculate the specific heat of the metal: c = Q/mcΔT = 462.4 J/(0.20 kg)(150oC - 40oC) = 6.3 J/g°C

4. If heat losses to the surroundings are not negligible, the answer will be smaller than the actual value for specific heat of the metal.

Question:

Given below are observations on molar specific heats at room temperature of some common gases Gas Molar specific heat (Cv​)(calmol^−1K^−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 ?

Answer:

Answer: The difference in molar specific heats of the common gases compared to monatomic gases can be explained by the fact that the common gases are composed of molecules with multiple atoms, while the monatomic gas is composed of single atoms. The larger number of atoms in the molecules of the common gases increases the amount of energy required to increase the temperature of the gas, resulting in a higher molar specific heat.

The higher molar specific heat value for chlorine can be attributed to the fact that chlorine is a diatomic gas, composed of two atoms. Since it is composed of two atoms, more energy is required to increase its temperature compared to other monatomic gases, resulting in a higher molar specific heat value.

Question:

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 1.250×10^5Pa 0.200×10^5Pa Normal melting point of sulphur 1.797×10^5Pa 0.287×10^5Pa (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 ?

Answer:

(a) Temperature of normal melting point of sulphur as read by thermometer A = 1.797×10^5Pa Temperature of normal melting point of sulphur as read by thermometer B = 0.287×10^5Pa

(b) The slight difference in answers of thermometers A and B is due to the fact that they are using different gases (oxygen and hydrogen respectively). To reduce the discrepancy between the two readings, a comparison should be made between the two thermometers using a gas that is common to both. This could be done by measuring the temperature of a known substance, such as the triple point of water, with both thermometers and comparing the results.

Question:

A steel tape 1m long is correctly calibrated for a temperature of 27.0oC. The length of a steel rod measured by this tape is found to be 63.0cm on a hot day when the temperature is 45.0oC. What is the actual length of the steel rod on that day? What is the length of the same steel road on a day when the temperature is 27.0oC? Coefficient of linear expansion of steel =1.20×10^−5K^−1.

Answer:

1. Actual length of the steel rod on the hot day: 63.0 cm + (45.0 - 27.0) x (1.20 x 10^-5 x 63.0) = 63.0036 cm

2. Length of the same steel rod on a day when the temperature is 27.0oC: 63.0036 cm - (45.0 - 27.0) x (1.20 x 10^-5 x 63.0036) = 63.0 cm

Question:

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.

Answer:

1. Calculate the difference in temperature between the initial and final temperatures of the first problem: 80∘C - 50∘C = 30∘C

2. Calculate the difference in temperature between the initial and final temperatures of the second problem: 60∘C - 30∘C = 30∘C

3. Since the difference in temperature between the two problems is the same, the time it takes to cool from 60∘C to 30∘C will be the same as the time it took to cool from 80∘C to 50∘C, which was 5 minutes. Therefore, it will take 5 minutes to cool from 60∘C to 30∘C.