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DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET PDF Download

Introductory Exercise 19.1

Take cice = 0.53 cal/g-°C, cwater = 1.0 cal/g-°C,(Lf)water = 80cal/g and(Lv)water = 529 cal/g unless given in the question,
Ques 1: In a container of negligible mass 140 g of ice initially at -15° C is added to 200 g of water that has a temperature of 40°C. If no heat is lost to the surroundings, what is the final temperature of the system and masses of water and ice in mixture?
Sol: Let mixture is water at θ°C (where 0°C< θ < 40°C)
Heat given by water = Heat taken by ice
∴  (200) (1) (40 - θ) = (140) (0.53) (15) + (140)(80)+(140)(l)(θ - 0)
Solving we get,
θ = -12.7°C
Since, θ < 0° C and we have assumed the mixture to be water whose temperature can't be less than 0°C.
Hence, mixture temperature θ = 0° C. Heat given by water in reaching upto 0° C is,
θ = (200) (1) (40 - 0) = 8000 cal.
Let m mass of ice melts by this heat, then 8000 = (140) (0.53) (15)+(m) (80)
Solving we get m = 86 g
∴  Mass of water = 200 + 86 = 286 g
Mass of ice = 140 - 86 = 54 g

Ques 2: The temperatures of equal masses of three different liquids A, Sand Care 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C and when S and C are mixed is 23°C. What would be the temperature when A and C are mixed?
Sol: 
A + B msA (16 - 12) = msB (19' - 16)
4sA = 3sB
B + C msB (23 - 19) = msC (28 - 23)
4s= 5sC ...(ii)
Solving these two equations, we get
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
A + C
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
Solving, we get
θ = 20.25° C

Ques 3: Equal masses of ice (at 0°C) and water are in contact. Find the temperature of water needed to just melt the complete ice.
Sol:
mL = ms (θ - 0°)
∴  DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET

Ques 4: A nuclear power plant generates 500 MW of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise by 10°C, what is the required flow rate in kg/s?
Sol:

 DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET 

Introductory Exercise 19.2

Ques 1: Suppose a liquid in a container is heated at the top rather than at the bottom. What is the main process by which the rest of the liquid becomes hot?
Sol: 
In convection, liquid is heated from the bottom.

Ques 2: The inner and outer surfaces of a hollow spherical shell o f inner radius 'a' and outer radius 'b' are maintained at temperatures T1 and T2 (< T1). The thermal conductivity of material of the shell is k. Find the rate of heat flow from inner to outer surface.
Sol:
Let us take an element at distance r from centre of thickness dr. Applying the formula of DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET or thermal resistance.
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
= thermal resistance of this element
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
Rate of heat flow,
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET

Ques 3: Show that the SI units of thermal conductivity are W/m-K.
Sol: 
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET

Ques 4: A carpenter builds an outer house wall wit h a layer of wood 2.0 cm thick on the outside and a layer of an insulation 3.5 cm thick as the inside wall surface. The wood has k = 0.08 W/m -K and the insulation has k = 0.01 W/m -K. The interior surface temperature is 19° C and the exterior surface temperature is -10° C.
Sol:
(a) H1 = H2
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
or
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
∴  DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
Solving, we get
θ= -8.1°C
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
= 7.7 W/m2

Ques 5: A pot with a steel bottom 1.2 cm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m2. The water inside the pot is at 100°C and 0.440 kg are evaporated every 5.0 minute. Find the temperature of the lower surface of the pot, which is in contact with the stove. Take Lv = 2.256 × 10J/kg and ksteel = 50.2 W/m-K.
Sol: 
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
= 105°C

Ques 6: A layer of ice o f thickness y is on the surface of a lake. The air is at a constant temperature -θ°C and the ice water interface is at 0°C. Show that the rate at which the thickness increases is given by,
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
where k is the thermal conductivity of the ice, L the latent heat of fusion and p is the density of the ice.
Sol:
See the extra points just before solved examples. Growth of ice on ponds. We have already derived that,
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
∴  DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET

Ques 7: The emissivity of tungsten is 0.4. A tungsten sphere with a radius of 4.0 cm is suspended within a large evacuated enclosure whose walls are at 300 K. What power input is required to maintain the sphere at a temperature of 3000 K if heat conduction along supports is neglected?
Take σ= 5.67 × 10-8 W/m2 -K4.
Sol:
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
= (0.4)(5.67 × 10-8)(4π)(4 × 10-2)2[3000)4 -(300)4]
= 3.7 × 104W

Ques 8: Find SI units of thermal resistance.
Sol:
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET
DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET 

The document DC Pandey Solutions: Calorimetry & Heat Transfer | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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FAQs on DC Pandey Solutions: Calorimetry & Heat Transfer - Physics Class 11 - NEET

1. What is calorimetry and how does it relate to heat transfer?
Ans. Calorimetry is the scientific measurement of heat transfer. It involves measuring changes in temperature in order to determine the amount of heat gained or lost during a chemical or physical process. Calorimetry is closely related to heat transfer as it helps in understanding how heat energy is transferred from one object to another.
2. What are the different types of calorimeters used in calorimetry experiments?
Ans. There are several types of calorimeters used in calorimetry experiments. Some common types include bomb calorimeters, constant pressure calorimeters, and constant volume calorimeters. Bomb calorimeters are used to measure the heat of combustion, constant pressure calorimeters are used to measure heat transfer at constant pressure, and constant volume calorimeters are used to measure heat transfer at constant volume.
3. How does calorimetry help in determining the specific heat capacity of a substance?
Ans. Calorimetry plays a crucial role in determining the specific heat capacity of a substance. By measuring the heat gained or lost by a substance and the corresponding change in temperature, the specific heat capacity can be calculated using the formula Q = mcΔT, where Q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
4. What are some practical applications of calorimetry and heat transfer in everyday life?
Ans. Calorimetry and heat transfer have various practical applications in everyday life. For example, they are used in cooking to determine the amount of heat required to cook food, in HVAC systems to calculate the amount of heat needed to warm or cool a room, and in the automotive industry to study engine performance and efficiency by measuring heat transfer during combustion.
5. How does insulation affect heat transfer in calorimetry experiments?
Ans. Insulation plays a crucial role in minimizing heat transfer during calorimetry experiments. By insulating the calorimeter, heat losses to the surroundings are reduced, allowing for more accurate measurements of heat transfer. Insulation materials such as Styrofoam or air gaps help in reducing conduction and convection heat transfer, ensuring that the measured heat transfer is primarily due to the process under investigation.
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