Class 11th

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.

0 Follower 10 Views

Contributor-Level 10

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

...more

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

less

0 0

New answer posted

6 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?

0 Follower 2 Views

Contributor-Level 10

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

0 0

New answer posted

6 months ago

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

...more

0 Follower 1 View

Contributor-Level 10

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

...more

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.

less

0 0

New answer posted

6 months ago

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?

0 Follower 1 View

Contributor-Level 10

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

0 0

New answer posted

6 months ago

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?

0 Follower 2 Views

Contributor-Level 10

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

...more

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

less

0 0

New answer posted

6 months ago

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.

0 Follower 2 Views

Contributor-Level 10

The maximum allowable stress for the structure, P = $10^{9}$ Pa

Depth of the ocean, d = 3 km = 3 $* 10^{3}$ m

Density of water $ρ$ = $10^{3}$ kg/ $m^{3}$

Acceleration due to gravity, g = 9.8 m/s

The pressure exerted because of sea water at the depth d = $ρ d g$ = $10^{3} *$ 3 $* 10^{3} * 9.8$ Pa

= 2.94 $* 10^{7}$ Pa

The maximum allowable stress is more than the pressure, hence the structure is suitable.

0 0

New answer posted

6 months ago

11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1 K^–1; heat of fusion of water = 335 J g^–1 ).

0 Follower 4 Views

Contributor-Level 10

11.13 Mass of the copper block, m = 2.5 kg = 2.5 $* 10^{3}$ gm

Rise in temperature of the copper block, $?$ T = 500°C

Specific heat of the copper, C = 0.39 g–1 K–1

Heat of fusion of water, L = 335 J g–1

The maximum heat the copper block can lose, Q = mc $?$ T = 2.5 $* 10^{3} * 0.39 * 500$ = 487500 J

Let $m_{1}$ gm be the mass of the ice, which will melt because of the copper block.

Heat gained by ice block = Q = $m_{1} L$

$m_{1} = \frac{Q}{L}$ = $\frac{487500}{335}$ g = 1455.22 gm = 1.45 kg

0 0

New answer posted

6 months ago

10.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^–3. Determine the height of the wine column for normal atmospheric pressure.

0 Follower 4 Views

Contributor-Level 10

Density of mercury, $ρ_{1}$ = 13.6 $* 10^{3}$ kg/ $m^{3}$

Density of wine, $ρ_{2}$ = 9.84 $* 10^{2}$ kg/ $m^{3}$

Height of the mercury column for atmospheric pressure, $h_{1}$ = 760 mm = 0.76 m

Height of the mercury column for atmospheric pressure = $h_{2}$

From the relation, P = $ρ g h$ , since the pressure on both the system are equal

$ρ_{1} g h_{1}$ = $ρ_{2} g h_{2}$ , we get $h_{2}$ = $\frac{ρ_{1} g h_{1}}{ρ_{2} g}$ = $\frac{13.6 * 10^{3} * 0.76}{9.84 * 10^{2}}$ = 10.5 m

0 0

New answer posted

6 months ago

11.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g^–1 K^–1.

0 Follower 2 Views

Contributor-Level 10

11.12 Power of the drilling machine, P = 10 kW= 10 $* 10^{3}$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $* 10^{3}$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $* 60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $δ$ T

Total energy consumed by the drilling machine= P $* t$ = 10 $* 10^{3} * 150$ J = 1.5 $* 10^{6}$ J

It is given 50% of energy is useful.

So useful energy, $? Q$ = 50% of Pt= 0.5 $*$ 1.5 $* 10^{6} =$ 7.5 $* 10^{5}$ J

We know, $? Q$ = mc $?$ T or T = $\frac{? Q}{m c}$ = $\frac{7.5 * 10^{5}}{8 * 10^{3} * 0.91}$ = 103 $?$

Therefore 2.5 minute drilli

...more

11.12 Power of the drilling machine, P = 10 kW= 10 $* 10^{3}$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $* 10^{3}$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $* 60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $δ$ T

Total energy consumed by the drilling machine= P $* t$ = 10 $* 10^{3} * 150$ J = 1.5 $* 10^{6}$ J

It is given 50% of energy is useful.

So useful energy, $? Q$ = 50% of Pt= 0.5 $*$ 1.5 $* 10^{6} =$ 7.5 $* 10^{5}$ J

We know, $? Q$ = mc $?$ T or T = $\frac{? Q}{m c}$ = $\frac{7.5 * 10^{5}}{8 * 10^{3} * 0.91}$ = 103 $?$

Therefore 2.5 minute drilling will raise the block temperature by 103 $?$

less

0 0

New answer posted

6 months ago

10.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

0 Follower 2 Views