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Question-1

Steam at 100° C is allowed to pass into a vessel containing 10 grams of ice and 100 grams of water at 0° C, until all the ice is melted and the temperature is raised to 5° C. Neglecting water equivalent of the vessel and the loss due to radiation etc., calculate how much steam is condensed.

Solution:
Let m be the mass of the condensed steam.

Heat lost by steam = m × 536 + m (100-5) = 631 m cal

Heat gained by water = 100 × 1 × 5 = 500 cal

Heat gained by ice = 10 × 80 + 10 × 5 = 850 cal

Heat lost = Heat gained

631 m ; = 500 + 850

.

Question-2

Specific heat of steam = 1 cal/gm
Latent heat of steam = 540 cal/gm.

 


Solution:
Let m be the mass of the steam required per hour to raise the temperature of water.

Heat gained by water in the boiler in one hour = 10,000 × 1 × (80 - 20)

= 600000 calories.
Heat lost by the steam = m × 1 × (150 - 100) + m (540) + m × 1 × (100 - 90)

= 50 m + 540 m + 10m
= 600 m calories

Heat lost = Heat gained
600 m = 600000
m = 1000 = 1 k gm

Thus 1 kg steam is required per hour.

Question-3

The temperature of equal masses of three different liquids A, B and C are 12° C, 19° C, 28° C respectively. The temperature when A and B are mixed is 16° C and when B and C are mixed is 23° C. What would be the temperature when A and C are mixed?

Solution:
Let m be the mass of each liquid. Again let the specific heat of the liquids be sA, sB and sC respectively.

(i) When A and B are mixed

Heat lost by B = m sB (19 – 16) = 3 m sB
Heat gained by A = m sA (16 – 12) = 4 m sA
Heat lost = Heat gained

3 m sB = 4 m sA

(ii) When B and C are mixed

Heat lost by C = m sC (28 – 23) = 5 m sC
Heat gained by B = m sB (23 – 19) = 4 m sB
4 sB = 5 sC

From the above two cases:

(iii) When A and C are mixed,
Let the temperature of the mixture be T.
Heat lost by C = m sC (28 – T)
Heat gained by A = m sA (T – 12)

Question-4

Latent heat of fusion of ice = 80 cal/gm
Latent heat of vaporization = 540 cal/gm

 


Solution:
Heat lost by 200 gm of steam when it is condensed to water at 100° C = 200 × 540 = 108000 cal

Heat gained by 200 gm of ice at 0° C = m L + m s T

= 200 × 80 + 200 × 1 × (100 – 0) = 36000 cal
Total heat gained = 36000 + 30000 = 66000 cal

We can notice that the amount of heat lost is greater than the amount of heat gained. Thus only one part of the steam will condense to water at 100° C. Let M be the mass of the steam condensed.

M × 540 = 66000 or M = 122.2 gm

Final temperature of the system will be 100° C.
Weight of the contents = 250 + 200 + 122.2 = 572.2 gm.

Question-5

How much steam at 100° C will just melt 3200 gm of ice at -10° C? (Specific heat of ice = 0.5 and latent heat of steam = 540 cals)

Solution:
Let m grams be the mass of the steam.
Heat lost by the steam = m × L + m × 1 × (100 – 0)
= m × 540 + 100 m = 640 m
Heat gained by ice = m × s × T + m L
= 3200 × 0.5 × [ 0 – (-10)] + 3200 × 80
= 27200 cal
640 m = 27100 or
m = 425 gm.

Question-6

Given,
Specific heat capacity of copper = 0.42 × 103 J kg-1 K-1
Specific heat capacity of water = 4.2 × 103 J kg-1 K-1
Latent heat of fusion of ice = 3.36 × 105 J kg-1
Latent heat of condensation of steam = 22.5 × 105 J kg-1

 


Solution:
Let the calorimeter originally contains x gm of ice and (200 – x) gm of water.

Heat lost by steam = m L + m s T = m [L + s T]

=

= 73800 joule

Heat gained by calorimeter = 100 / 1000 = (0.42 × 103) × (50 – 0)

= 2100 joule

Heat gained by water =

= (42000 – 210x) joule

Heat gained by ice = m L + m s T = m [L + sT]

=

= 546x joule

Heat lost = Heat gained

73800 = 2100 + (42000 – 210x) + 546x

x = 88.4 gm
Mass of ice in the original mixture = 200 – 88.4 = 111.6 gm
Ratio of ice present to water present = 88.4 : 111.6 = 1 : 1.26.

Question-7

Coefficient of volume expansion of water = 1.5 × 10-4 (° C)-1
Density of mercury = 13.6 gm cm-3
Density of water = 1 gm cm-3
Specific heat of mercury = 0.03 cal gm-1 (° C)-1

 


Solution:
Let after the addition of heat, the temperature of water and mercury rises to t ° C. Now

Heat gained by water + heat gained by mercury = Total heat supplied

500 × 1 × t + 1000 × 0.03 × t = 21200
530 t = 21200
t = 40° C

Supposing that the given values of densities are for the initial temperature, then initial volume of water = = 500 cm3

Increase in volume of water = 500 × γ ω × t
= 500 × (1.5 × 10-4) × 40 = 3 cm3

Density of water at final temperature =

Volume of water over flown = 3.52 × 1.006 = 3.541 cm3
Increase in volume of Hg = 3.541 – 3 = 0.541 cm3

So,

Question-8

Given that,
sAl = 0.22 cal g-1 ° C-1, ρ Al = 2.7 g cm-3
sCu = 0.1 cal g-1 ° C-1, ρ Cu = 8.9 g cm-3

 


Solution:
Let m be the mass of the cylinder. Then heat lost by aluminium cylinder

Q1 = m × sAl × 50
Similarly, heat lost by copper cylinder
Q2 = m × sCu × 50

Let rAl and rCu be the radii of aluminium and copper cylinders respectively.

Then m = π (rAl)2hρ Al = π (rCu)2hρ Cu

Let hAl and hCu be the depths of penetration of aluminium and copper cylinders respectively. Now heat received by the first ice block

Q1 = π (rAl)2 hAlρQ2 = π (rCu)2 hCuρ

Where ρ is the density of ice and L is its latent heat

Now we have


Question-9

 


Solution:
The portion AB represents the change of state from solid state to liquid state at constant temperature or the portion AB represents the latent heat of fusion. The portion CD represents the change of state from liquid state to vapour state at constant temperature or the portion CD represents the latent heat of vaporization.

(ii) CD = 2 AB
i.e., Latent heat of vaporization is twice the latent heat of fusion.

(iii) The slope DE is equal to dT/dQ for vapour state, i.e., this gives the rate of increase of temperature of vapour with heat input.

Slope of DE
Specific heat of vapour

(iv) Slope OA > Slope BC
The slope OA represents that
Specific heat of solid

Now slope OA > slope BC, represents that specific heat of the liquid state is more than that of the solid state.

Question-10

A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc is 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350 K? Assume, for purpose of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Solution:
The rate of flow of heat through the block temperature 400 K and T is

This heat is taken by the disc. The rate at which the heat is taken by disc is given by

As the system is thermally insulated

Integrating the above expression, we get

Question-11

A large steel wheel is to be fitted on to a shaft of the same material. At 27° C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. AT what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range.

Solution:
Coefficient of linear expansion of steel = 1.20 × 10-5 K-1

Here T1 = 27° C = 27 + 273 = 300 K
Length at temperature, T1K = lT1 = 8.70 cm
Length of temperature, T2K = lT2 = 8.69 cm
Change in length = lT2 – lT1 =lT1 α (T2 – T1)
8.69 – 8.70 = 8.70 × (1.20 × 10-5) × (T2 – 300)

T2 – 300 =

T2 = 300 – 95.8 = 204.2 K = -68.8° C
Thus the temperature of the shaft = -68.8° C
.

Question-12

A liquid takes 5 minutes to cool from 80° C to 50° C. How much time will it take to cool from 60° C to 30° C? The temperature of surrounding is 20° C.

Solution:
According to Newton’s law,

(Excess of temperature over surroundings)

If T be the excess of temperature over surroundings

Where m is the mass of a body of specific heat s and it cools through a small range of temperature dT. Negative sign is used to indicate that there is a gall of temperature

On Integration, we get logeT = - Kt

If T1 be the temperature excess over surrounding at time t1 and T2 at time t2 then

Where t is the time elapsed.

Or 2.3026[log10T1 – log10T2] = Kt

In first case, T1 = (80° C – 20° C) = 60° C, T2 = (50° C - 20° C) = 30° C and t = 5 min.

2.3026[log1060 – log1030] = 5K

In second case, T1 = (60° C - 20° C) = 40° C, T2 = (30° C - 20° C) = 10° C and t =?

2.3026[log1040 – log1010] = Kt

Dividing the equations of the two cases, we get

Question-13

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.

Solution:
Here fall in temperature = Δ T = 101° F - 98° F = 3° F = 3 × (5/9) ° C = (5/3) ° C

Mass of the child m = 30 kg

Specific heat of human body = Specific heat of water

C = 1000 cal kg-1 ° C-1

Heat lost by the child, Δ Q = mcΔ T = 30 × 1000 × (5/3) = 50000 cals

Here m’ be the mass of water evaporated in 20 minutes then, m’ L = Δ Q

Average rate of extra evaporation = (86.2 / 20) = 4.31 g min-1.

Question-14

Consider a compound slab, consisting of two materials having different thickness, L1 and L2, and different thermal conductivities, k1 and k2. If the temperatures of the outer surfaces are T2 and T1, find the rate of heat transfer through the compound slab in a steady state.

Solution:
Let Tx be the temperature at the interface between the two materials. Then,

In steady state Δ Q1/Δ t = Δ Q2/Δ t, so that

Let Δ Q/Δ t be the rate of heat transfer (the same for all sections). Then, solving for Tx and substituting into either of these equations, we obtain

The extension to any number of sections in series is given by

.

Question-15

A 75 gm block of copper, taken from a furnace, is dropped into a 300-gm glass beaker containing 200-gm of water. The temperature of the water rises from 12° C to 27° C. What was the temperature of the furnace?

Solution:
This is an example of two systems originally at different temperatures reaching thermal equilibrium after contact. No mechanical energy is involved, only heat exchange. Hence,

Heat lost by copper = Heat gained by (beaker + water)

i.e., mCcC (TC – Te) = (mGcG + mWcW) (Te – TW)

The subscript C stands for copper, G for glass, and W for water. The initial copper temperature is TC, the initial beaker water temperature is TW. And Te is the equilibrium temperature.

Substituting the given values, with cC = 0.093 cal/gm° C, cG = 0.12 cal/gm° C, and cW = 1.0 cal/gm° C, we get

(75 gm) (0.093 cal/gm° C) (TC - 27° C) = [(300gm) (0.12 cal/gm° C) + (200 gm) (1.0 cal/gm° C)] (27° C - 12° C)

TC = 530° C.





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