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 T_{1} and T_{2} (< T_{1}). 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/mK.
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) H_{1} = H_{2}
∴
or
∴
∴
Solving, we get
θ= 8.1°C
= 7.7 W/m^{2}
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 m^{2}. 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 × 10^{6 }J/kg and k_{steel} = 50.2 W/mK.
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/m^{2} K^{4}.
Sol:
= (0.4)(5.67 × 10^{8})(4π)(4 × 10^{2})2[3000)^{4} (300)^{4}]
= 3.7 × 10^{4}W
Ques 8: Find SI units of thermal resistance.
Sol:
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