Grade XII Physics Chapter 11 heat problems Sindh board
Karachi and Sindh Grade XII, 2nd years, class 12, HSSC Part-2 Fsc Physics Notes Chapter 11 Heat questions answers, and Problems.
Grade XII Physics Chapter 11 heat Notes
Table of Contents
Q.1) How do you distinguish between temperature and heat? Give examples.
Temperature is a quantitative property possessed by a material. It helps us to recognize which body is hot and which one is hotter or colder. For example, the body which has a low temperature when it is placed in contact with some heat source will have a rise in temperature with the passage of time. This can be felt by touching it.
Heat on the other hand is the amount of energy that is transferred from one body to another because of differences in their temperatures. For example, when we place a hotter and colder body in contact with each other then after some time the temperature of both bodies will be the same. This is due to temperature transfer. The temperature will transfer from a hotter body to a colder body until both bodies will be in thermal equilibrium.
Hence we can say that temperature is the characteristic of a body while heat is a property dependent on temperature.
Q.2) Why is the earth not In thermal equilibrium with the sun?
Thermal equilibrium is a state obtained by placing two bodies in contact with each other.
As in the case of the sun and earth, the earth is not directly in contact with the sun, therefore, the temperature of the earth is not the same as that of the sun.
Earth is in contact with the surrounding environment and hence the temperature of the earth is in thermal equilibrium with the environment. Sun comprises a very small portion of the environment, therefore it does not have a thermal equilibrium with the earth.
If the sun comprises that maximum part of the surrounding earth, then the earth will be in thermal equilibrium with the sun.
Q.3) Is temperature a microscopic or macroscopic concept?
Temperature is a macroscopic property as it is a measurement of heat.
It is the emission of all radiation from a body.
Q.4) It is observed that when a mercury In a glass thermometer is put In a flame, the column of mercury first descends and then rises. Explain?
When we put a mercury thermometer in flame then the column of mercury first descends and then rises. This is because at first, the capillary tube containing mercury got in contact with heat. Due to this heat, the capillary tube expands which results in a drop of mercury level in the thermometer.
When the capillary tube reaches its maximum expansion then the heat will transfer to the mercury inside it. Then due to heat absorption, the mercury will expand and there will be a rise in the level of mercury in the thermometer.
Q.5) Is it correct that unit for specific heat capacity Is m2 S -2 (Co ) -1
SI unit of specific heat capacity is joule per kilogram per kelvin. We know that,
J Kg-1 K-1 = (Kgm2s-2) (Kg-1) (K-1)
=m2s-2K-1 = m2s-2 (Co + 273.1)-1
Therefore we can say that a unit of specific heat capacity can be joule per kilogram per kelvin.
Q.6) What is the standard temperature?
Standard temperature is defined as zero degrees Celsius, which translates to 32 degrees Fahrenheit and 273.15 degrees kelvin. This temperature is necessary for freezing water in the air at standard pressure.
Q.7) When a block with a hole in It Is heated, why does not the material around the hole expand into the hole and make it small?
When we heat the box with a hole in it then the whole does not get smaller due to the expansion of the box but it gets wider.
When we heat the block then all of its dimensions will expand linearly including the dimensions of the hole. It can be visualized as when the material of the box expands then its molecular bonds are lengthened.
Consider two identical boxes without any holes. Now cut a hole in one of the boxes and then heat up both the boxes. Both of the blocks will expand and so will the hole. That is the reason that the hole will also expand when the box expands.
Q.8) A thermometer is placed in direct sunlight. Will it read the temperature of the air, or of the sun. or of something else.
When we place a thermometer in sunlight then it will read the temperature of the surrounding air. As the thermometer reads the temperature of the material with which it is connected directly.
Therefore we can say that the thermometer will read the temperature of the air surrounding the thermometer. If the thermometer was in direct contact with the sun then it would read the temperature of the sun itself.
Q.9) Will kilograms of hydrogen contain more atoms than kilograms of lead? Explain.
Yes, one kilogram of hydrogen contains more atoms than one kilogram of lead.
As hydrogen atoms are much lighter as compared to lead atoms, therefore we contain more hydrogen atoms in a given mass than any other material.
Q.10) The pressure in a gas cylinder containing hydrogen will leak more quickly than if it is containing oxygen. Why?
If we had two gas cylinders, one containing oxygen in it and the other containing hydrogen. Then the pressure in a gas cylinder containing hydrogen will leak more quickly than the cylinder containing oxygen.
This is due to the fact that hydrogen is a lighter gas or liquid s compared to oxygen and lighter liquids flow more quickly as compared to heavier.
This can be understood by the example of two containers containing water and honey. When there is a leak point in both of the cylinders then the pressure of the cylinder containing water will release more quickly as compared to the cylinder containing honey.
Q.11) What are some factors that affect the efficiency of automobile engines?
The efficiency of an automobile engine depends on,
Temperature of hot reservoir
Temperature of cold reservoir
Therefore, these will be the factors which can affect the efficiency of an automobile engine.
Q.12) What happens to the temperature of a room in which an air conditioner is left running on a table in the middle of the room.
The operating function of an air condition works in a way that it extracts heat from one side and releases it to the other side. If we have an air conditioner in the middle of the room on a table then it will extract heat from its front side and will release it to the opposite side. Therefore, there will be no overall change in the temperature of the room as heat extracted and released is in the same room.
That is the reason that we fix our air conditioners on a wall such that it can release the heat outside the room. In that way we can have our room cold.
Q.13) When a sealed thermos bottle full of hot coffee is shaken, what are the changes, if any in
a) the temperature of the coffee and
b) The Internal energy of the coffee.
When we shake the bottle then the collision of the molecules of coffee will increase and as a result of that increase in collision,
Temperature of coffee will increase and
Internal energy of coffee will also increase
Problems Sindh G12 Physics Chapter 11 (Heat)
Q.1) i. The normal body temperature Is 98.4° F . What is this temperature on Celsius scale ? ii. At what temperature do the Fahrenheit and Celsius
Answer: i) Given Data: Temperature of normal body = TF = 98.4oF To Find: Temperature in Celsius = TC = ? Solution: We know that ToC = (5/9) (ToF – 32) Putting values, we get TC = (0.56) (98.4 – 32) = 36.88oC Tc=36.88oC………………………………..Answer ii) We know that ToC = (5/9) (ToF – 32) Consider, TC = TF = X Then putting this, we get X = (5/9) (X – 32) = (5/9)X – 17.78 Or, X – (5/9)X = 17.78 X ( 1-5/9) = 17.78 X (0.444) = 17.78 Therefore, x=40o………………………………………….Answer
Q.2) A steel rod has a length of exactly 0.2 cm at 30°C. What will be its length at 60°C
Answer: Given Data: Length of steel rod = L1 = 0.2 cm Temperature = T1 = 30oC = 303o K Temperature = T2 = 60oC = 333o K Temperature co efficient = α = 11 × 10-6K-1 To Find: Length at 60oC = L2 = ? Solution: We know that, L2 = L1 (1 + α ΔT) = L1 [1 + α (T2 – T1)] Putting values, we get L2 = (0.2cm) [1 + 11 × 10-6K-1 (333oK – 303oK)] = (0.2cm) [1 + 11 × 10-6K-1 (30oK)] L2 = (0.2cm) [1 + 11 × 10-6K-1 (30oK)] = 0.20066 cm L2=0.20066 cm……………………………..Answer
Q.3) Find the change in volume of an aluminium sphere of 0.4m radius when it is heated from 0 to 100°C.
Answer: Given Data: Radius of sphere = r = 0.4 m Initial Temperature = Ti = 0oC = 273oK Final Temperature = Tf = 100oC = 373oK Temperature co efficient for aluminium = α = 24 × 10-6K-1 To Find: Change in volume = ΔV = ? Solution: We know that, ΔV = βV1ΔT ……………………………… (1) Where, β = 3α = 3 × 24 × 10-6K-1 β = 72 × 10-6K-1 And,
And, ΔT = Tf – Ti = 373oK – 273oK = 100oK Putting values of β, V1 and ΔT in equation (1), we get ΔV = (72 × 10-6K-1) (0.268m3) (100oK) = 0.0019 m3 ΔV =0.0019 m3…………………………………..Answer
Q.4) Calculate the root mean square speed of hydrogen molecule at 800K.
Answer: Given Data: Temperature = T = 800K To Find: Root man square value of hydrogen = Vrms = ? Solution: We know that,
We know that,
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Putting values, we get
Answer: Vrms= 3158 ms-1
Chapter 11 | XII Physics | Sindh Textbook Board
Q.5) a. Determine the average value of the kinetic energy of the particles of an ideal gas at 0 oC and at 500 C.
b) What is the kinetic energy per mole of an ideal gas at these temperatures ?
Answer: a) Given Data: i) First temperature = T = 0oC = 0oC + 273 = 273 k ii) Second temperature = T = 50oC = 50 + 273 = 323K To Find: Kinetic energy at 0oC = K.E = ? Kinetic energy at 50oC = ? Solution: i) We know that,
Putting values, we get
K.E = 5.65 x 10-21 j……………………..Answer ii) We Know that,
Putting values, we get
K.E = 6.68 x 10-21 j…………………Answer b) To Fine:
Kinetic Energy per molecule at 0oC = ?
Kinetic Energy per molecule at 50oC = ?
We know that kinetic energy per molecule is given by,
Putting values, we get
K.E per molecule = 3402.9 J/mol………………………….Answer ii) We know that kinetic energy per molecule is given by,
Putting values, we get
i.e K.E per molecule = 4026.2 J/mol………………………..Answer
Q.6) A 2 kg iron block is taken from a furnace where its temperature was 650°C and placed on a large block of ice at 0°C. Assuming that all the heat given up by the iron is used to melt the ice, how much ice is melted.
Answer: Given Data: Mass of iron = m1 = 2 kg Temperature of furnace = T1 = 650oC Temperature of ice = T2 = 0oC To Find: Amount of ice melted = m2 = ? Solution: We know that according to the law of heat exchange we have, Heat lost by iron block = Heat gained by the ice i.e. m1C1ΔT = m2Hf Which gives,
Where, C1 is the specific heat capacity of iron and is given by, C1 = 499.8 J kg-1K-1 And Hf is the latent heat of fusion of ice and is given by, Hf = 336000 J kg-1 And, ΔT = T2 – T1 = 650oK Putting values in equation (1), we get
Answer: m2 = 1.93 Kg
Q.7) In a certain process 400 J of heat are supplied to a system and at the sametime 150J of work are done by the system. What is the increase in the internal energy of the system. (250J)
Answer: Given Data: Heat supplied = ΔQ = 400 J Work Done = ΔW = 150 J To Find: Increase in internal energy = ΔU = ? Solution: We know that, ΔU = ΔQ – ΔW Putting values, we get ΔU = 400J – 150J = 250 J ΔU = 250J……………………………..Answer
Q.8) There Is an increase of internal energy by 400 Joules when 800 Joules of work is done by a system. What is the amount of heat supplied during this process?
Answer: Given Data: Increase in internal energy = ΔU = 400 J Work Done = ΔW = 800 J To Find: Heat supplied = ΔQ = ? Solution: We know that, ΔU = ΔQ – ΔW ⇒ ΔQ = ΔU + ΔW Putting values, we get ΔQ = 400J + 800J = 1200 J ΔQ = 1200J…………………………. Answer
Q.9) A heat engine performs 200J of work in each cycle and has efficiency of 20 percent. For each cycle of operation a) How much heat is absorbed and b) How much heat is expelled?
Answer: Given Data: Work Done = ΔW = 200 J Efficiency of engine = E = 20 % To Fine:
Heat absorbed = Q1 = ?
Heat expelled = Q2 = ?
Solution: a) We know that,
Q1 = 1000 J………………………………..Answer b) We know that, ΔW = Q1 – Q2 ⇒ Q2 = Q1 – ΔW Putting values, we get Q2 = 1000J – 200J = 800 J Q1 = 800 J…………………………………….Answer
Q.10) A heat engine operates between two reservoirs at temperatures of25°C and 300°C. What is the maximum efficiency for this engine?
Answer: Given Data: Temperature of first reservoir = T1 = 300oC = 300 + 273 = 573K Temperature of second reservoir = T2 = 25oC = 25 + 273 = 298K To Find: Maximum efficiency = E = ? Solution: We know that maximum efficiency is given by,
Putting values, we get
Answer: E = 48%
Q.11) The low temperature reservoir of a carnot engine is at 7°C and has an efficiency of 40%. It is desired to increase the efficiency to 50%. By how much degrees the temperature of hot reservoir be increased.
Answer: Given Data: Temperature of low temperature reservoir = T2 = 7oC = 7 + 273 = 280K First efficiency = E1 = 40% Increased efficiency = E2 = 50% To Find: Increase in temperature of hot reservoir = ΔT = ? Solution: Efficiency of cold reservoir is given by,
Putting value of E1, we get T1 (100 – 40) = 28000K T1 = 466.67K Similarly efficiency of increased temperature is given by,
Putting value of E1, we get T1’ (100 – 50) = 28000K T1’ = 560K Then we know that, increase in temperature is given by, ΔT = T1’ – T1 = 560K – 466.7 k ΔT = 93.4 K……………………………….Answer