Class 11th

11.18 A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^–1 m^–1 K^–1 ; Heat of vaporization of water = 2256 * 10³J kg^–1.

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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:

$θ$ = $\frac{K A (T_{1} - T_{2}) t}{l}$ , where

$T_{1}$ = Flame temperature in contact with the boiler

$T_{2}$ = Boiling point of water = 100 $?$

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

By equating for $θ$ we get

6 $* 2256 * 10^{3}$ = 109&n

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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:

$θ$ = $\frac{K A (T_{1} - T_{2}) t}{l}$ , where

$T_{1}$ = Flame temperature in contact with the boiler

$T_{2}$ = Boiling point of water = 100 $?$

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

By equating for $θ$ we get

6 $* 2256 * 10^{3}$ = 109 $* 0.15 * (T_{1} - T_{2}) * 60 / 0.01$

$(T_{1} - T_{2})$ = 137.98 $?$

$T_{1}$ = 237.98 $? (F l a m e t e m p e r a t u r e i n c o n t a c t w i t h t h e b o i l e r)$

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8 months ago

10.11 Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.

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No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river. In rapid, the flow is turbulent whereas Bernoulli's equation is applicable for laminar flow only.

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8 months ago

11.17 A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 * 103 J kg–1]

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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 $* 60 * 60$ 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 $* 10^{3}$ J ${k g}^{- 1}$

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

The amount of heat lost by the food: $θ$ = $\frac{K A (T - 0) t}{l}$ , where

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

$θ$ = $\frac{0.01 * 0.54 (45 - 0) * 6 * 60 * 60}{0.05}$ = 104976 J

We also know $θ = m' L$ so m' = 104976/ (335 $* 10^{3})$ = 0.313 kg

Hence the amount of ice remains after 6 h = 4

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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 $* 60 * 60$ 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 $* 10^{3}$ J ${k g}^{- 1}$

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

The amount of heat lost by the food: $θ$ = $\frac{K A (T - 0) t}{l}$ , where

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

$θ$ = $\frac{0.01 * 0.54 (45 - 0) * 6 * 60 * 60}{0.05}$ = 104976 J

We also know $θ = m' L$ so m' = 104976/ (335 $* 10^{3})$ = 0.313 kg

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

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8 months ago

10.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)

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Height of the spirit column, $h_{1}$ = 12.5 + 15 cm = 27.5 cm

Height of the water column, $h_{2}$ = 10 + 15 cm = 25 cm

Density of spirit, $ρ_{1}$ = 0.8 gm/ ${c m}^{3}$

Density of water, $ρ_{2}$ = 1 gm/ ${c m}^{3}$

Density of mercury, $ρ$ = 13.6 gm/ ${c m}^{3}$

Let h be the difference between the levels of mercury in two limbs

Pressure exerted by mercury column of h height= h $ρ g$ = 13.6hg …. (i)

Difference between pressure exerted by water and spirit columns:

= $h_{2} ρ_{2} g - h_{1} ρ_{1} g = g (25 * 1 - 27.5 * 0.8) =$ 3g ……. (ii)

Equating equations (i) and (ii), we get

13.6hg = 3g

h = 3/13.6 = 0.221 cm

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8 months ago

11.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g ^–1.

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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 = $T_{1} - T_{2} = (101 ° F -$ 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 $? θ = m c ? T$ = 30 $* 1000 * 1.666$ = 49980 cal

Let $m_{1}$ be the mass of water evaporated from the child's body in 20 mins.

Loss of heat $? θ$ = $m_{1} * L$ = 580 $m_{1}$ = $49980 c a l$

$m_{1} =$ &nb

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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 = $T_{1} - T_{2} = (101 ° F -$ 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 $? θ = m c ? T$ = 30 $* 1000 * 1.666$ = 49980 cal

Let $m_{1}$ be the mass of water evaporated from the child's body in 20 mins.

Loss of heat $? θ$ = $m_{1} * L$ = 580 $m_{1}$ = $49980 c a l$

$m_{1} =$ 86.2 gm

Rate of evaporation = $\frac{m_{1}}{t}$ = 4.31 g/min

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8 months ago

10.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

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Height of the spirit column, $h_{1}$ = 12.5 cm = 0.125 m

Height of the water column, $h_{2}$ = 10 cm = 0.1m

$P_{0}$ = atmospheric pressure

$ρ_{1}$ = density of spirit, $ρ_{2}$ = density of water

Pressure at point B = $P_{0}$ + $h_{1} ρ_{1} g$

Pressure at point D = $P_{0}$ + $h_{2} ρ_{2} g$

But pressure at point B and D are same. Hence

$P_{0}$ + $h_{1} ρ_{1} g$ = $P_{0}$ + $h_{2} ρ_{2} g$ or $h_{1} ρ_{1}$ = $h_{2} ρ_{2}$

$\frac{ρ_{1}}{ρ_{2}}$ = $\frac{h_{2}}{h_{1}}$ = $\frac{10}{12.5}$ = 0.8

So the specific gravity of spirit is 0.8

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11.15 Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat (Cv )

(cal mo1^–1 K^–1)

Hydrogen &n

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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

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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.

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10.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?

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The maximum mass of the car can be lifted, m = 3000 kg

Area of cross-section of the load carrying piston, A = 425 ${c m}^{2}$ = 425 $* 10^{- 4} m^{2}$

Maximum force exerted by the load, F = mg = 3000 $* 9.8$ N = 29400 N

Maximum pressure exerted, P = F/A = (29400 / 425 $* 10^{- 4}$ ) Pa = 6.917 $* 10^{5}$ Pa

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11.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

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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 $* 1 = 150 g$

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, $θ$ = mC $?$ T ……. (i)

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

Heat gained by the water and calor

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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 $* 1 = 150 g$

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, $θ$ = mC $?$ T ……. (i)

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

Heat gained by the water and calorimeter system; $θ'$ = (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 = $\frac{(M + m') C_{w} ? T'}{m ? T}$ = $\frac{(150 + 25) * 4.186 * 13}{200 * 110}$ = 0.433 J/g/ $°$ K

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10.7 A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

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