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

Introductory Exercise 19.1

Take c_ice = 0.53 cal/g-°C, c_water = 1.0 cal/g-°C,(L_f)_water = 80cal/g and(L_v)_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 ms_A (16 - 12) = ms_B (19' - 16)
4s_A = 3s_B
B + C ms_B (23 - 19) = ms_C (28 - 23)
4s_B = 5s_C ...(ii)
Solving these two equations, we get

A + C

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°)
∴  

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:
  

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  or thermal resistance.

= thermal resistance of this element

Rate of heat flow,


Ques 3: Show that the SI units of thermal conductivity are W/m-K.
Sol: 
∴
 

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

∴
or

∴
∴  
Solving, we get
θ= -8.1°C


= 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 L_v = 2.256 × 10⁶ J/kg and k_steel = 50.2 W/m-K.
Sol: 

∴

= 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,

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,

∴
∴  

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:
= (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: