28: Heat Transfer : Neet-xii-physics

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NEET-XII-Physics

28: Heat Transfer

with Solutions - page 3

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Qstn #6
A solid sphere and a hollow sphere of the same material and of equal radii are heated to the same temperature.
(a) Both will emit equal amount of radiation per unit time in the biginning
(b) Both will absorb equal amount of radiation from the surrounding in the biginning.
(c) The initial rate of cooling (dT/dt) will be the same for the two spheres
(d) The twp spheres will have equal temperature at any instant
digAnsr: a,b
Ans : (a) Both will emit equal amount of radiation per unit time in the beginning.
(b) Both will absorb equal amount of radiation from the surrounding in the beginning.
Let the temperature of the surroundings be `` {T}_{0}``.
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
`` u=\sigma A{T}^{4}``
Here, `` \sigma `` is Stephen's constant.
Also, the energy absorbed per unit time by the body is given by
`` {u}_{0}=e\sigma A{{T}_{0}}^{4}``
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.
Page No 98:

#
Section : iv

Qstn #1
A uniform slab of dimension 10 cm × 10 cm × 1 cm is kept between two heat reservoirs at temperatures 10°C and 90°C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W m^-1 °C^-1. Find the amount of heat flowing through the slab per minute.
Ans : Given:
Thermal conductivity of the material, k = 0.80 W m^-1 °C^-1
Area of the cross section of the slab, A = 100 cm² = 10^-2 m²
Thickness of the slab, Δx = 1 cm = 10^-2 m
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta T}{{\displaystyle \frac{\,\mathrm{\,\Delta \,}x}{k·A}}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\left(90-10\right)k·\,\mathrm{\,A\,}}{\,\mathrm{\,\Delta \,}x}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\left(80\right)\times 0.8\times {10}^{-2}}{{10}^{-2}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=64\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=64\times 60=3840\,\mathrm{\,J\,}/\,\mathrm{\,min\,}``
Page No 98:

Qstn #2
A liquid-nitrogen container is made of a 1 cm thick styrofoam sheet having thermal conductivity 0.025 J s^-1 m^-1 °C^-1. Liquid nitrogen at 80 K is kept in it. A total area of 0.80 m² is in contact with the liquid nitrogen. The atmospheric temperature us 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.
Ans : `` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
Thickness of the container, l = 1 cm = 10^{`` -``2} m
Thermal conductivity of the styrofoam sheet, `` k=0.025\,\mathrm{\,J\,}{\,\mathrm{\,s\,}}^{-1}{\,\mathrm{\,m\,}}^{-1}°{\,\mathrm{\,C\,}}^{-1}``
Area, A = 0.80 m²
Thermal resistance, `` \frac{l}{kA}=\frac{{10}^{-2}}{{\displaystyle 0.025\times 0.80}}``
Temperature difference, `` ∆T={T}_{1}-{T}_{2}=300-80=220\,\mathrm{\,K\,}``
Rate of flow of heat, `` \left(\frac{\Delta Q}{\Delta t}\right)=\frac{{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=\frac{220}{{10}^{-2}}\times 0.025\times 0.80``
`` \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=440\,\mathrm{\,J\,}/s``
Page No 98:

Qstn #3
The normal body-temperature of a person is 97°F. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature = 47°F, surface of the body under clothes = 1.6 m², conductivity of the cloth = 0.04 J s^-1 m^-1°C^-1, thickness of the cloth = 0.5 cm.
Ans : `` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
Temperature of the body, `` {T}_{1}=97°\,\mathrm{\,F\,}=36.11°\,\mathrm{\,C\,}``
Temperature of the surroundings, `` {T}_{2}=47°\,\mathrm{\,F\,}=8.33°\,\mathrm{\,C\,}``
Conductivity of the cloth, `` K=0.04\,\mathrm{\,J\,}{\,\mathrm{\,s\,}}^{-1}{\,\mathrm{\,m\,}}^{-1}°{\,\mathrm{\,C\,}}^{-1}``
Thickness of the cloth, `` l=0.5\,\mathrm{\,cm\,}=0.005\,\mathrm{\,m\,}``
Area of the cloth, `` A=1.6{\,\mathrm{\,m\,}}^{2}``
Difference in the temperature, `` ∆T={T}_{1}-{T}_{2}=36.11-8.33=27.78°\,\mathrm{\,C\,}``
Thermal resistance = `` \frac{l}{KA}=\frac{0.005}{\left(0.04\right)\times 1.6}`` = 0.078125
Rate at which heat is flowing out is given by
`` \frac{\Delta Q}{\Delta t}=\frac{{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{KA}}}``
`` \frac{\Delta Q}{\Delta t}=\frac{27.78}{0.078125}``
`` \frac{\Delta Q}{\Delta t}=356\,\mathrm{\,J\,}/s``
Page No 98:

Qstn #4
Water is boiled in a container having a bottom of surface area 25 cm², thickness 1.0 mm and thermal conductivity 50 W m^-1°C^-1. 100 g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporisation of water = 2.26 × 10⁶ J kg^-1.
Ans : Area of the bottom of the container, `` A=25{\,\mathrm{\,cm\,}}^{2}=25\times {10}^{-4}{\,\mathrm{\,m\,}}^{2}``
Thickness of the bottom of the container, l = 1 mm = 10^{`` -``3} m
Latent heat of vaporisation of water, `` L=2.26\times {10}^{6}\,\mathrm{\,J\,}{\,\mathrm{\,kg\,}}^{-1}``
Thermal conductivity of the container, `` k=50\,\mathrm{\,W\,}{\,\mathrm{\,m\,}}^{-1}°{\,\mathrm{\,C\,}}^{-1}``
mass = 100 g = 0.1 kg
Rate of heat transfer from the base of the container is given by
`` \frac{\Delta Q}{\Delta t}=\frac{mL}{\Delta t}=\frac{\left(0.1\right)\times 2.26\times {10}^{6}}{1\,\mathrm{\,min\,}}``
`` \frac{\Delta Q}{\Delta t}=0.376\times {10}^{4}\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
Also,
`` \frac{\Delta Q}{\Delta t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow 0.376\times {10}^{4}=\frac{T-100}{{\displaystyle \frac{{10}^{-3}}{50\times 25\times 10{}^{-4}}}}``
`` \Rightarrow 0.376\times {10}^{4}=\frac{50\times 25\times {10}^{-4}\left(T-100\right)}{{10}^{-3}}``
`` \Rightarrow \left(T-100\right)=3.008\times 10``
`` \Rightarrow T=130°\,\mathrm{\,C\,}``
Page No 98:

Qstn #5
One end of a steel rod (K = 46 J s^-1 m^-1°C^-1) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm². Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 10⁵ J kg^-1.
Ans :
Given:
Thermal conductivity of the rod, K = 46 J s^-1 m^-1 °C^-1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm²
= 0.04 × 10^-4 m²
= 4 × 10^-6 m²
Rate of transfer of heat is given by
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta T}{{\displaystyle \frac{l}{\,\mathrm{\,kA\,}}}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\left({T}_{1}-{T}_{2}\right)kA}{l}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\left(\frac{100-0}{1}\right)\times 46\times 4\times {10}^{-6}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=184\times {10}^{-4}\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
`` \,\mathrm{\,Also\,},\,\mathrm{\,\Delta Q\,}=m{\,\mathrm{\,L\,}}_{f}``
`` \therefore \left(\frac{m}{t}\right){L}_{f}=184\times {10}^{-4}``
`` \Rightarrow \left(\frac{m}{t}\right)\times 3.36\times {10}^{5}=184\times {10}^{-4}``
`` \Rightarrow m=\frac{184\times {10}^{-4}}{3.36\times {10}^{5}}\times t``
`` \Rightarrow m=5.5\times 10\times {10}^{-9}\times 1\,\mathrm{\,kg\,}/\,\mathrm{\,s\,}``
`` \Rightarrow m=5.5\times {10}^{-8}\times {10}^{3}\,\mathrm{\,g\,}/\,\mathrm{\,s\,}``
`` \Rightarrow m=5.5\times {10}^{-5}\,\mathrm{\,g\,}/\,\mathrm{\,s\,}``
Page No 98:

Qstn #6
A icebox almost completely filled with ice at 0°C is dipped into a large volume of water at 20°C. The box has walls of surface area 2400 cm², thickness 2.0 mm and thermal conductivity 0.06 W m^-1°C^-1. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 10⁵ J kg^-1.
Ans : Area of the walls of the box, A = 2400 cm² = 2400 × 10^-4 m²
Thickness of the ice box, l = 2 mm = 2 × 10^-3 m
Thermal conductivity of the material of the box, K = 0.06 W m^-1 °C^-1
Temperature of the water outside the box, T₁ = 20°C
Temperature of ice, T₂= 0°C
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{{T}_{1}-{T}_{2}}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\left(\frac{20}{2\times {10}^{-3}}\right)\times 0.06\times 2400\times {10}^{-4}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=24\times 6=144\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
Rate at which the ice melts =`` \frac{m{L}_{f}}{t}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\left(\frac{m}{t}\right){L}_{f}``
`` \Rightarrow 144=\left(\frac{m}{t}\right)\times 3.4\times {10}^{5}``
`` \Rightarrow \frac{m}{t}=\frac{144}{3.4\times {10}^{5}}\,\mathrm{\,kg\,}/\,\mathrm{\,s\,}``
`` \Rightarrow \frac{m}{t}=\frac{144\times 60\times 60}{3.4\times {10}^{5}}\,\mathrm{\,kg\,}/\,\mathrm{\,h\,}``
`` \Rightarrow \frac{m}{t}=1.52\,\mathrm{\,kg\,}/\,\mathrm{\,h\,}``
Page No 98:

Qstn #7
A pitcher with 1-mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of 0.1 g s^-1. The surface area of the pitcher (one side) = 200 cm². The room temperature = 42°C, latent heat of vaporization = 2.27 × 10⁶J kg^-1, and the thermal conductivity of the porous walls = 0.80 J s^-1 m^-1°C^-1. Calculate the temperature of water in the pitcher when it attains a constant value.
Ans : Thickness of porous walls, l = 1 mm = 10^-3 m
Mass, m = 10 kg
Latent heat of vapourisation, L_v = 2.27 × 10⁶ J/kg
Thermal conductivity, K = 0.80 J/m s °C
ΔQ = 2.27 × 10⁶ × 10 J
0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 10⁵ s.
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{2.27\times 10{}^{7}}{{10}^{5}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=2.27\times {10}^{2}\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow \frac{\Delta \,\mathrm{\,Q\,}}{\,\mathrm{\,\Delta \,}t}=\left(\frac{42-T}{{10}^{-3}}\right)·0.80\times 2\times {10}^{-2}``
`` \Rightarrow 2.27\times {10}^{2}=\frac{\left(42-\,\mathrm{\,T\,}\right)}{{10}^{-3}}\times 0.80\times 2\times {10}^{-2}``
`` \Rightarrow T=27.8°\,\mathrm{\,C\,}``
`` \Rightarrow T=28°\,\mathrm{\,C\,}``
Page No 98:

Qstn #8
A steel frame (K = 45 W m^-1°C^-1) of total length 60 cm and cross sectional area 0.20 cm², forms three sides of a square. The free ends are maintained at 20°C and 40°C. Find the rate of heat flow through a cross section of the frame.
Ans : Thermal conductivity, K = 45 W m^-1 °C^-1
Length, l = 60 cm = 0.6 m
Area of cross section, A = 0.2 cm² = 0.2 × 10^-4 m²
Initial temperature, T₁ = 40°C
Final temperature, T₂ = 20°C
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temerature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{kA({T}_{1}-{T}_{2})}{{\displaystyle l}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{45\times 0.2\times {10}^{-4}\left(40-20\right)}{0.6}``
`` =0.03\,\mathrm{\,W\,}``
Page No 98:

Qstn #9
Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm². The walls of the vessel are adiabatic but the flat parts are made of 1-mm thick aluminium (K = 200 J s^-1 m^-1°C^-1). Assume that the outside temperature is 20°C. The density of water is 100 kg m^-3, and the specific heat capacity of water = 4200 J k^-1 g °C^-1. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.
Ans : Area of cross section, A = 10 cm² = 10 × 10^-4 m²
Thermal conductivity, K = 200 Js^-1 m^-1 °C^-1
Height, H = 10 cm
Length, l = 1 mm =10^-3 m
Temperature inside the cylindrical vessel, T₁ = 50°C
Temperature outside the vessel, T₂ = 30°C

Rate of flow of heat from 1 flat surface will be given by
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{{T}_{1}-{T}_{2}}{{\displaystyle \frac{l}{\,\mathrm{\,kA\,}}}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\left(50-30\right)\times 200\times {10}^{-3}}{{10}^{-3}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=6000\,\mathrm{\,J\,}/\,\mathrm{\,sec\,}``
Heat escapes out from both the flat surfaces.
Net rate of heat flow = 2 × 6000 = 12000 J/sec
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{m·\,\mathrm{\,s\,}·\Delta T}{{\displaystyle \,\mathrm{\,\Delta \,}t}}``
`` ``
Mass = Volume density
⇒ `` {10}^{-3}\times 0.1\times 1000``
`` =0.1\,\mathrm{\,g\,}``
Using this in the above formula for finding the rate of flow of heat, we get
`` 12000=0.1\times 4200\times \left(\frac{\Delta T}{t}\right)``
`` \Rightarrow \frac{\,\mathrm{\,\Delta \,}T}{\,\mathrm{\,\Delta \,}t}=\frac{12000}{420}=28.57``
`` \,\mathrm{\,As\,}\,\mathrm{\,\Delta \,}T=1°\,\mathrm{\,C\,}``
`` \Rightarrow \frac{1}{t}=28.57``
`` \Rightarrow t=\frac{1}{28.57}=0.035\,\mathrm{\,sec\,}``
Page No 98:

Qstn #10
The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm²) is maintained at 20°C and the right end is maintained at 80°C. Neglecting any loss of heat through radiation, find
a) the temperature at a point 11 cm from the left end and
(b) the heat current through the rod. Thermal conductivity of copper = 385 W m^-1°C^-1.
Ans :
Area of cross section, A = 0.2 cm² = 0.2 × 10^-4 m²
Thermal conductivity, k = 385 W m^-1 °C^-1
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{kA\mathit{(}{T}_{1}-{T}_{2}\mathit{)}}{{\displaystyle l}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\left(\frac{80-20}{0.2}\right)\times 385\times 0.2\times {10}^{-4}``
`` =2310\times {10}^{-3}``
`` =2.31\,\mathrm{\,J\,}/\,\mathrm{\,sec\,}``
Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\,\mathrm{\,k\,}A\,\mathrm{\,\Delta \,}T}{{\displaystyle l}}``
`` \Rightarrow \frac{\,\mathrm{\,\Delta \,}T}{l}=\left(\frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}\right)\times \frac{1}{\,\mathrm{\,kA\,}}``
`` \frac{T-20}{11\times {10}^{-2}}=\frac{2.31}{383\times 0.2\times {10}^{-4}}``
`` \Rightarrow T=33+20``
`` \Rightarrow T=53°\,\mathrm{\,C\,}``
Page No 99:

Qstn #11
The ends of a metre stick are maintained at 100°C and 0°C. One end of a rod is maintained at 25°C. Where should its other end be touched on the metre stick so that there is no heat current in the rod in steady state?
Ans :
One end of the rod is at a temperature of 25°C. So, if no heat current flows through the rod in steady state, then the other end of the rod should also be at a temperature of 25°C.
Let the point at which the other end of the rod is touched be C.
No heat flows through the rod when the temperature at point C is also 25°C.
Heat current through AC = Heat current through CB
`` \Rightarrow \frac{{\left(\,\mathrm{\,\Delta \,}T\right)}_{\,\mathrm{\,AC\,}}}{{\displaystyle \frac{x}{\,\mathrm{\,kA\,}}}}=\frac{{\left(\,\mathrm{\,\Delta \,}T\right)}_{\,\mathrm{\,CB\,}}}{{\displaystyle \frac{100-x}{\,\mathrm{\,kA\,}}}}``
`` \frac{\left(100-25\right)}{x}=\frac{25-0}{100-x}``
`` \frac{3}{x}=\frac{1}{100-x}``
`` 300-3x=x``
`` 300=4x``
`` x=75\,\mathrm{\,cm\,}``
Thus, it should be touched at 75 cm from 100°C end.
Page No 99:

Qstn #12
A cubical box of volume 216 cm³ is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5°C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.
Ans :
Volume of the cube, V = a³ = 216 cm³
Edge of the cube, a = 6 cm
Surface area of the cube = 6a²
= 6 (6 × 10^-2)²
= 216 × 10^-4 m²
Thickness, l = 0.1 cm = 0.1 × 10^-2 m
Temperature difference, Δ T = 5°C
The inner surface of the cube is heated by a 100 W heater.
∴ Power, P = 100 W
Power = Energy per unit time
∴ Rate of flow of heat inside the cube, R = 100 J/s
Rate of flow of heat is given by
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta \,\mathrm{\,T\,}}{{\displaystyle \frac{l}{\,\mathrm{\,kA\,}}}}``
`` 100=\frac{\,\mathrm{\,k\,}\times 216\times {10}^{-4}\times 5}{0.1\times {10}^{-2}}``
`` \,\mathrm{\,k\,}=0.9259\,\mathrm{\,W\,}/\,\mathrm{\,m\,}°\,\mathrm{\,C\,}``
Page No 99:

Qstn #13
Figure (28-E1) shows water in a container having 2.0 mm thick walls made of a material of thermal conductivity 0.50 W m^-1°C^-1. The container is kept in a melting-ice bath at 0°C. The total surface area in contact with water is 0.05 m². A wheel is clamped inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady state is reached in which the block goes down with a constant speed of 10 cm s^-1 and the temperature of the water remains constant at 1.0°C. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in contact. Take g = 10 m s^-2.
Figure
Ans :
Temperature of water, T₁ = 1°C
Temperature if ice bath, T₂ = 0°C
Thermal conductivity, K = 0.5 W/m °C
Length through which heat is lost, l = 2 mm = 2 × 10^-3 m
Area of cross section, A = 5 × 10^-2 m²
Velocity of the block, v = 10 cm/sec = 0.1 m/s
Let the mass of the block be m.
Power = F · v
= (mg) v ......(i)
Also,
`` \,\mathrm{\,Power\,}=\frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}.......\left(\,\mathrm{\,ii\,}\right)``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}\,\mathrm{\,t\,}}=\frac{\,\mathrm{\,k\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\,\mathrm{\,l\,}}........\left(\,\mathrm{\,iii\,}\right)``
`` ``
From equation (i), (ii) and (iii), we get
`` \left(\mathit{m}\mathit{g}\right)v=\frac{\,\mathrm{\,k\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{l}``
`` m=\frac{0.5\times 5\times {10}^{-2}\left(1\right)}{\left(2\times {10}^{-3}\right)\times 10\times 0.1}``
`` m=12.5\,\mathrm{\,kg\,}``
Page No 99: