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Speed of the wind on the upper surface of the wing, $V_1$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_2$ = 63 m/s

Area of the wing, A = 2.5 $m^2$

Density of air, $\rho$ = 1.3 kg/ $m^3$

According to Bernoulli’s theorem, we have the relation:

$P_1$ + $\frac{1}{2}\rho {V_1}^2$ = $P_2$ + $\frac{1}{2}\rho {V_2}^2$

( $P_2$ - $P_1$ )= $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ , where $P_1$ = pressure on the upper surface of the wing and $P_2$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_2$ - $P_1$ )A = $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ A = $\frac{1}{2}\times 1.3\times ({70}^2$ - ${63}^2)\times 2.5$ N = 1512.8 N = 1.5 $\times {10}^3$ N

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02m

Glycerine mass flow rate, M = 4 $\times {10}^{-3}$ kg/s

Density of glycerine, $\rho$ = 1.3 $\times {10}^3$ kg

Viscosity of glycerine, $\eta$ = 0.83 Pa-s

Now, volume of glycerine flowing per sec V = $\frac{M}{\rho }$ = $\frac{4\times {10}^{-3}}{1.3\mathrm{}\times {10}^3}$ $m^3$ /s = 3.08 $\times {10}^{-6}$ $m^3$ /s

According to Poiseville’s formula, we know the flow rate

V = $\frac{\pi \times p\times r^4}{8\times \eta \times l}$ where p is the pressure difference between two ends of the tube

p = $\frac{V\times 8\times \eta \times l}{\pi \times r^4}$ = $\frac{3.08\mathrm{}\times {10}^{-6}\times 8\times 0.83\times 1.5}{\pi \times {\left(0.01\right)}^4}$ = 976.47 Pa

Reynolds’s number is given by the relation

Re = $\frac{4\rho V}{\pi d\eta }$ = $\frac{4\times 1.3\mathrm{}\times {10}^3\times 3.08\mathrm{}\times {10}^{-6}}{3.1416\times 0.02\times 0.83}$ = 0.3

Since the Reynolds’s number is 0.3, the flow is laminar

11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.

(b) Brass is a good conductor of heat, when one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a colder value and one feels cold.

On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.

Thus a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Black body radiation equation is given by:

E = $\sigma$ ( $T^4$ - $T_o^4$ ), where

E = energy radiation

$\sigma$ = constant

$T=$ temperature of the optical pyrometer

$T_o$ = Temperature of open space

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}$

When the same piece of iron placed in a furnace, the radiation energy is E = $\sigma T^4$

(d) In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from Earth’s surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat ( 540 cal/g)

11.18 Base area of the boiler, A = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Let us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s–1 m–1 K–1

The amount of heat flowing into water through the brass base of the boiler is given by:

$\theta$ = $\frac{KA\left(T_1-T_2\right)t}{l}$ , where

$T_1$ = Flame temperature in contact with the boiler

$T_2$ = Boiling point of water = 100 $?$

Heat required for boiling water $\theta$ = mL, where L = heat of vaporization of water = 2256 × 103 J kg–1

By equating for $\theta$ we get

6 $\times 2256\times {10}^3$ = 109 $\times 0.15\times \left(T_1-T_2\right)\times 60/0.01$

$\left(T_1-T_2\right)$ = 137.98 $?$

$T_1$ = 237.98 $?\left(\mathrm{F}\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{b}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{r}\right)$

11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m

Thickness of the icebox, l = 5 cm = 0.05 m

Mass of ice kept in the box, m = 4 kg

Time gap, t = 6 h = 6 $\times 60\times 60$ s

Outside temperature, T = 45 °C

Coefficient of thermal conductivity of thermocole, K = 0.01 J $s^{-1}{m}^{-1}{K}^{-1}$

Heat of fusion of water, L = 335 $\times {10}^3$ J ${kg}^{-1}$

Let m’ be the mass of the ice melts in 6 h

The amount of heat lost by the food: $\theta$ = $\frac{KA\left(T-0\right)t}{l}$ , where

A = Surface area of the box = 6 $s^2$ = 6 $\times {0.3}^2{m}^2$ = 0.54 $m^2$

$\theta$ = $\frac{0.01\times 0.54\left(45-0\right)\times 6\mathrm{}\times 60\times 60}{0.05}$ = 104976 J

We also know $\theta =m\text{'}L$ so m’ = 104976/ (335 $\times {10}^3)$ = 0.313 kg

Hence the amount of ice remains after 6 h = 4 – 0.313 = 3.687 kg

11.16 Initial body temp of the child,  $T_1$ = 101°F

Final body temp of the child,  $T_2$ = 98°F

Change in temperature,  $?$ T = $T_1-T_2=(101°\mathrm{F}-\mathrm{}$ 98°F) = 3 °F = $\frac{5}{9}$  (3-32) $°$ C = 1.666 $?$

Time taken t achieve this temperature, t = 20 min

Specific heat of human body = Specific heat of water, c = 1000 cal/kg/ $?$

Latent heat of evaporation of water, L = 580 cal/g

Mass of the child, m = 30 kg

The heat lost by the child is given as $?\theta =mc?\mathrm{T}$ = 30 $\times 1000\times 1.666$ = 49980 cal

Let $m_1$ be the mass of water evaporated from the child’s body in 20 mins.

Loss of heat $?\theta$ = $m_1\times L$ = 580 $m_1$ = $49980cal$

$m_1=$ 86.2 gm

Rate of evaporation = $\frac{m_1}{t}$ = 4.31 g/min

11.15 The gases listed above are diatomic. Besides the translational degree of freedom, they have other degrees of freedom. Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence the molar specific heat of diatomic gases is more than that of monatomic gases.

If only rotational mode of motion considered, then the molar specific heat of a diatomic gas

= $\frac{5}{2}$ R = $\frac{5}{2}*1.98$ = 4.95 cal mo1^–1 K^–1

With the exception of Chlorine, all the observations given above agrees with ( $\frac{5}{2}$ R). This is because at room temperature, chlorine also has vibrational modes besides rotational and translational modes of motion.

11.14 Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal,  $T_1$ = 150 °C, Final temperature of the metal,  $T_2$ = 40 °C

The water equivalent mass of the calorimeter, m’ = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass of water, M at T = 27 °C, = 150 $\times 1=150g$

Fall in metal temperature,  $?$ T = $T_1-T_2$ = 150 – 40 = 110 °C

Specific heat of water,  $C_w$ = 4.186 J/g/ $°$ K

Let the specific heat of metal = C

Then, heat loss by the metal,  $\theta$ = mC $?$ T ……. (i)

Rise in the water of the calorimeter system $?$ T’ = 40 – 27 = 13°C

Heat gained by the water and calorimeter system; $\theta \text{'}$ = (M + m’) $C_w?$ T’ …… (ii)

Now heat lost by the metal = heat gained by the water and calorimeter system

mC $?$ T = (M + m’) $C_w?$ T’

C = $\frac{(\mathrm{M}\mathrm{}+\mathrm{}\mathrm{m}\mathrm{’})C_w?\mathrm{T}\mathrm{’}}{m?\mathrm{T}}$ = $\frac{(150+25)\times 4.186\times 13}{200\times 110}$ = 0.433 J/g/ $°$ K