Contents
Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.
Apparent Loss in Weight of a Solid Immersed in a Liquid, at Different Temperatures
There is an apparent loss in the weight of a solid when it is immersed in a liquid. This loss in weight is due to the upthrust it receives in a liquid. This upthrust depends directly on the density of the liquid as well as the volume of the immersed portion of the body. With the change in temperature, density of the liquid and volume of the solid, both change. Hence, the apparent weight of a solid will be different at different temperatures.
Let the weight of a solid in air = W.
Weight of the solid completely immersed in a liquid at temperature t1 = W1 and that at temperature t2 = W2.
Hence, the apparent loss in weight at temperature t1 = W – W1 = M1g, and that at t2 = W – W2 = M2g, where M1 and M2 are the masses of the liquid displaced by the body. If V1 and ρ1 are the volume and density at t1, and V2 and ρ2 are those at t2 respectively, then,
M1 = V1ρ1 and M2 = V2ρ2
If the coefficient of real expansion of the liquid is γ and the coefficient of volume expansion of the material of the solid is γs, then,
Hence, the apparent weight of a body immersed in a liquid increases with the increase in temperature of the liquid.
Numerical Examples
Example 1.
A piece of metal weighs 50 g × g in air. It weighs 45 g × g when immersed in a liquid at 25°C, and 45.1 g × g at 100°C. If the coefficient of linear expansion of the metal is 12 × 10-6°C-1, find the coefficient of real expansion of the liquid.
Solution:
Apparent loss in weight at 25°C
= M1g = (50 – 45)g × g = 5 g × g
Apparent loss in weight at 100°C
= M2g = (50 – 45.1)g × g = 4.9g × g
As M2 = M1[1 – (γ – γs)(t2 – t1)]
∴ 4.9 = 5[1 – (γ – 12 × 10-6 × 3)(100 – 25)]
or, 4.9 = 5[1 – (γ – 12 × 10-6 × 3) × 75] [as γs = 3 × αs]
or, (γ – 36 × 10-6) × 75 × 5 = 5 – 4.9
or, γ – 36 × 10-6 = \(\frac{0.1}{75 \times 5}\)
or, γ = 36 × 10-6 + 2.67 × 10-4 = 3.03 × 10-4°C-1.
Example 2.
A glass rod weighs 90 g × g in air. It weighs 49.6 g × g when immersed in a liquid at 12°C, and 51.9 g × g at 97°C. Find the real expansion coefficient of the liquid. [Volume expansion coefficient of glass = 2.4 × 10-5°C-1]
Solution:
Let the volume of the glass rod be V1 at 12°C, and density of the liquid be ρ1.
The mass of the displaced liquid at that temperature
= 90 – 49.6 = 40.4 g
∴ Volume of the glass rod at that temperature,
V1 = \(\frac{40.4}{\rho_1}\) …… (1)
Again at 97°C, mass of the displaced liquid
= 90 – 51.9 = 38.1 g
Let at 97°C the volume of the glass rod be V2 and density of the liquid be ρ2.
Example 3.
Apparent weights of a solid in a liquid are 50 g × g and 52 g × g at 25°C and 75°C respectively. If the coefficient of linear expansion of the solid is αs = 6.6 × 10-6°C-1 and γ for the liquid is 7.3 × 10-4°C-1, what is the real weight of the solid in air?
Solution:
Let the real weight of the solid in air = M g × g.
Apparent loss in weight at 25°C = M1g = (M – 50)g × g
and apparent loss in weight at 75°C = M2g = (M – 52)g × g
t = t2 – t1 = 50°C, γs = 3αs = 3 × 6.6 × 10-6°C-1.
Now, M2 = M1[1 – (γ – γs) × t]
∴ (M – 52) = (M – 50)[1 – (7.3 × 10-4 – 19.8 × 10-6) × 50]
= (M – 50)[1 – 7.102 × 10-4 × 50]
= (M – 50) × 0.96449
or, M = 0.96449 M – 48.2245 + 52
or 0.03551M = 3.7755
or, M = \(\frac{3.7755}{0.03551}\) = 106.32 g
∴ Real weight of the solid is 106.32 g × g.
Example 4.
A sphere of mass 266.5 g and of diameter 7 cm floats on a liquid. When the liquid is heated to 35°C the sphere starts sinking In the liquid. If the density of the liquid at 0°C is 1.527 g ᐧ cm-3, find its coefficient of volume expansion. Neglect the expansion of the sphere.
Solution:
Volume of the sphere = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3\) cm3
∴ Volume of displaced liquid at 35°C = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3\) cm3
If the density of the liquid at 35°C is ρ35, then the mass of the displaced liquid at 35°C = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3\) × ρ35g.
From the condition of floatation,
Example 5.
A piece of metal weighs 46 g × g in air. It weighs 30g × g in a liquid of specific gravity 1.24 at 27°C. At 42°C, when the specific gravity of the liquid is 1.20, the weight of the piece immersed in it is 30.5 g × g. Find the coefficient of linear expansion (α) of the metal.
Solution:
Mass of the displaced liquid at 27°C
= 46 – 30 = 16g
Volume of the displaced liquid at 27°C, V27 = \(\frac{16}{1.24}\) cm3
Similarly volume of the displaced liquid at 42°C,
V42 = \(\frac{46-30.5}{1.20}\) = \(\frac{15.5}{1.20}\)cm3
So the volume of the piece at 27°C and 42°C are \(\frac{16}{1.24}\)cm3 and \(\frac{15.5}{1.20}\)cm3 respectively.
Now, V42 = V27{1 + γ(42 – 27)}
[where γ = coefficient of volume expansion of the metal]
Anomalous Expansion of Water
When a liquid is heated, its volume increases and density decreases with rise in temperature. Exceptions are observed in case of water for a certain range of temperature. When heated from 0°C to 4°C, the volume of water decreases and the density increases. Above 4°C, the volume of water increases again with the increase in temperature. Hence, water has a maximum density and a minimum volume at 4°C. Also, on cooling from 4°C to 0°C, the volume of water increases instead of decreasing. This exceptional behaviour of water in respect of expansion within the range of 0°C to 4°C, is called anomalous expansion of water.
The two graphs below represent the change in volume and density of 1 g of water, with the increase in temperature [Fig.]
Conclusions from the two graphs are:
i) Volume of water decreases as its temperature rises from 0°C to 4°C. Hence, expansion of water is anomalous and does not follow the general rule. Consequently, the coefficient of volume expansion of water is negative for this temperature range.
ii) Water has the least volume and the maximum density at 4°C.
iii) After 4°C, expansion of water takes place following the general rule. This means that, with the increase in temperature, the volume also starts increasing. The expansion is no longer anomalous.
iv) The slope of the curves for volume or for density, at temperature close to 4°C is almost zero. So, there is practically no change in volume or density for a small variation of temperature at and around 4°C. hence, the density of water at 4°C is taken as unity.
Effect of Anomalous Expansion of Water on Marine Life
Due to anomalous expansion of water, fishes and various living creatures can survive under frozen lakes, rivers or seas.
In cold countries, with the fall in atmospheric temperature, upper surfaces of lakes, seas and various ponds graduallý cool. Water of the upper surface, then being denser and heavier, moves down. Water below it, being comparatively warmer and lighter, moves up. This convection process in water continues until the density of the water in the lower part becomes maximum i.e., the temperature of the lower water reaches 4°C.
As the temperature of the upper surface decreases further below 4°C, density begins to decrease. So water cannot move down further. It then begins to cool further and at last turns into ice. As ice is lighter than water, a thick layer of ice, thus formed, floats over the surface of water. Both ice and water are bad conductors of heat. So, a negligible amount of heat can be conducted from the lower levels of water to the atmosphere outside. So, the entire volume of water (top to bottom) in a pond cannot freeze.
The temperature just below the floating ice-slab remains at 0°C [Fig.]. The temperature of water gradually increases with depth and the lowest layer remains at 4°C, where the living creatures can easily survive.
Numerical Examples
Example 1.
The area of cross-section of the capillary tube of a mercury thermometer is A0 and the volume of the bulb filled with mercury is V0 at 0°C. Find the length of the mercury column In the capillary tube as the bulb is heated to t°C. Coefficient of linear expansion of glass is α and coefficient of volume expansion of mercury is β.
Solution:
Let Vg and Vm be the volumes of the bulb and mercury at t°C.
∴ Vg = V0(1 + γgt) = V0(1 + 3αt)
and Vm = V0(1 + βt)
Hence, volume of mercury entering the tube is
Vm – Vg = V0(1 + βt) – V0(1 + 3αt) = V0t(β – 3α)
If the area of cross-section of the capillary tube at t°C is
then At = A0(1 + 2αt).
If the length of mercury column in the tube is 1,
then At × l = V0t(β – 3α)
[neglecting αβ and α2 as they are very small]
Both the equations (1) and (2), indicate the length of mercury column in the capillary tube at t°C.
Example 2.
A 1 L flask contains some mercury. It is observed that the volume of air in the flask remains unchanged at all temperatures. What is the volume of mercury in the flask? Coefficient of linear expansion of the material of the flask = 9 × 10-6°C-1 and coefficient of real expansion of mercury = 1.8 × 10-4°C-1
Solution:
Let the volume of mercury be V and the rise in temperature be θ°C. When the volume of air in the flask remains unaltered, expansion of mercury = volume expansion of the flask.
∴ V × θ × 1.8 × 10-4 = 1000 × θ × 3 × 9 × 10-6
or, V = \(\frac{27 \times 10^{-3}}{1.8 \times 10^{-4}}\) = 150 cm3.
Example 3.
A barometer with a brass scale reads 75.34 cm at 25°C. The scale was graduated at 20°C. What will be the reading of the barometer at 0°C? Coefficient of linear expansion α for brass 18 × 10-6°C-1 and coefficient of linear expansion γ for mercury = 18 × 10-5°C-1.
Solution:
Due to expansion of the scale, the actual barometric reading will be greater than the apparent reading. Let the real reading be h and the apparent reading be H.
∴ h = H(1 + αt) = 75.34(1 + 18 × 10-6 × 5)
= 75.34(1 + 9 × 10-5)
or, h = 75.34 × 1.00009 cm
Again, mercury expands with increase in temperature.
Hence, the barometric reading at 0°C will be less than that at 20°C. Let at 0°C, the barometric reading be H0.
∴ H0 = h(1 – γt) = 75.34 × 1.00009 [1 – 18 × 10-5 × 25]
= 75.34 × 1.00009[1 – 0.0045]
= 75.34 × 1.00009 × 0.9955 = 75.01 cm.
Example 4.
Coefficients of volume expansion of benzene and wood are 1.2 × 10-3°C and 1.5 × 10-4°C-1 respectively. Their respective densities at 0°C are 900 kg ᐧ m-3 and 880 kg ᐧ m-3. Find the temperature at which wood will just immerse in benzene.
Solution:
The wood will just immerse in benzene at a temperature for which the densities are equal. Let the required temperature be t°C, when the density of both is ρ.
∴ 900 = ρ[1 + 1.2 × 10-3 × t] ……… (1)
and 800 = ρ[1 + 1.5 × 10-4 × t] …… (2)
Dividing (1) by (2) we get,
\(\frac{900}{880}\) = \(\frac{1+1.2 \times 10^{-3} t}{1+1.5 \times 10^{-4} t}\)
or, t(900 × 1.5 × 10-4 – 880 × 1.2 × 10-3) = 880 – 900
or, t = 21.7°C.
Example 5.
A metal piece of density 8 g ᐧ m-3 is suspended from a wooden hook by an weightless string. The tension
in the string is 56 g × g. What will be the tension in the string. If the system is immersed in a liquid at 40°C? The surrounding temperature during the experiment is 20°C. At 20°C the specific gravity of the liquid is 1.24. The coefficients of volume expansion of the liquid and the metal are 4 × 10-5°C-1 and 8 × 10-4°C-1 respectively.
Solution:
Volume of the metal piece at 20°C,
V20 = \(\frac{56}{8}\) = 7 cm3
∴ Volume at 40°C,
V40 = V20[1 + 8 × 10-4 × 20]
= 7(1 + 8 × 10-4 × 20)cm3
Volume of the displaced liquid = V40
∴ Mass of displaced liquid = V40 × ρ40 × g
∴ Upthrust
= V40 × ρ40 × g
= 7[1 + 8 × 10-4 × 20] × \(\frac{1.24}{1+20 \times 4 \times 10^{-5}}\) × 980
= 7 × 1.24[1 + 0.0161] [1 + 8 × 10-4]-1 × 980
= 7 × 1.24 × 1.016[1 – 0.0008] × 980
= 7 × 1.24 × 1.016 × 0.9992 × 980 = 8.81 × 980 dyn.
∴ Tension in the string
= (56 – 8.81) × 980 = 4.625 × 104 dyn.
Example 6.
A body, at 4°C, floats with 0.98 part of its volume immersed In water. At what temperature the body will just immerse in water? Coefficient of real expansion of water = 3.3 × 10-4°C-1. Neglect expansion of the solid body.
Solution:
Let the required temperature be t°C, and the volume of the body = V.
Let the densities of water at 4°C and t°C be d1 and d2 respectively.
∴ From the condition of floatation, V × 0.98 × d1 = V × d2
Example 7.
A solid at 0°C floats with 98% of its volume immersed in a liquid. The solid floats completely immersed when the temperature is raised to 25°C. If the coefficient of volume expansion of the solid is 2.6 × 10-6°C-1, find the coefficient of real expansion of the liquid.
Solution:
Let at 0°C,volume of the solid = V0, density of the liquid = ρ0; at 25°C, volume of the solid = V’ and density of the liquid = ρ’
Now, V’ = V0[1 + 2.6 × 10-6 × 25]
and ρ0 = ρ’[1 + γ × 25]
where γ = coefficient of real expansion of the liquid.
From the condition of floatation,
V0 × 0.98 × ρ0 = V’ρ’
or, V0 × 0.98 × ρ'[1 + γ × 25] = V0[1 + 2.6 × 10-6 × 25] × ρ’
or, 1 + 25γ = \(\frac{1+0.000065}{0.98}\) or, 25γ = 1.02047 – 1
∴ γ = 8.19 × 10-4°C-1.
Example 8.
A mercury thermometer contains 0.4 cm3 of mercury at 0°C. The diameter of the capillary tube of the thermometer is 0.2 mm. What should be the length of the scale to measure temperatures between 0°C to 100°C? The coefficient of apparent expansion of mercury = 1.7 × 10-4°C-1.
Solution:
The apparent expansion of mercury due to increase in temperature from 0°C to 100°C
= 0.4 × 1.7 × 10-4 × 1oo = 0.0068 cm3
The cross-section of the capillary tube = π(0.01)2 cm2
∴ The length of the temperature measuring scale
= \(\frac{0.0068}{\pi(0.01)^2}\) = 2.16 cm.
Example 9.
A and B are two thermometers made of glass and both contain the same liquid. Both thermometers have spherical bulbs. The Internal diameter of the bulb of A is 7.5 mm and radius of the capillary tube is 1.25 mm. The corresponding values for B are 6.2 mm and 0.9 mm. Find the ratio of the lengths between two consecutive graduations in thermometers A and B.
Solution:
Let the separations between two consecutive graduations for 10 in thermometers A and B be x cm and y cm respectively.
Hence, for the thermometer A, volume expansion of liquid in the bulb due to an increase of 1°C in temperature = vol
ume of x cm length in the tube,
\(\frac{4}{3} \pi\left(\frac{0.75}{2}\right)^3\) × γ’ × 1 = x × π(0.125)2 ……. (1)
where γ’ is the coefficient of apparent expansion of the liquid.
Similarly for the thermometer B,
\(\frac{4}{3} \pi\left(\frac{0.62}{2}\right)^3\) × γ’ × 1 = y × π × (0.09)2 …….. (2)
Example 10.
A container is filled up to the brim with 500 g of water and 1000 g of mercury. When 21200 cal of heat is supplied to the system, 3.52 g of water flows out of the container. Neglecting the expansion of the container, find the coefficient of real expansion of mercury. Given, volume expansion coefficient of water = 1.5 × 10-4°C-1, density of mercury = 13.6 gᐧ cm-3,density of water 1 g ᐧ cm-3 and specific heat capacity of mercury = 0.03 cal ᐧ g-1 ᐧ °C-1.
Solution:
Due to the application of heat, 3.52 g i.e., 3.52 cm3 of water flows out of the container.
∴ The total expansion of mercury and water = 3.52 cm3.
Let the rise in temperature = t°C.
Heat absorbed by mercury + heat absorbed by water 21200
∴ 500 × 1 × t+ 1000 × 0.03 × t = 21200 or, t = 40°C
∴ Expansion of water = 500 × 1.5 × 10-4 × 40 = 3 cm3
∴ Expansion of mercury = (3.52 – 3) = 0.52 cm3
∴ \(\frac{1000}{13.6}\) × γ × 40 = 0.52
[γ = volume expansion coefficient of mercury]
∴ γ = 1.768 × 10-4°C-1