Question

The specific heat of a substance at its boiling point or melting point
0
infinity
negative
between 0 and 1
10 Mass of gas is 300 gm and its specific heat at constant volume is 750 J/kg K. if gas is heated through 75°C at constant pressure of 105 N/{m^2}, it expands by volume
0.08×106cm3
. find
CP/CV
.
1.4
1.374
1.474
1.5
11 A solid ball with a mass m = 0.53 kg floats in a tank of water. The ball is made of material with a density of 400
kg/m3
. The density of water is 1000
kg/m3
. What fraction of the volume of the ball is below the waterline?
0.1
0.2
0.3
0.4
12 A slab of wood with mass m = 1.7 kg floats 78% submerged.The density of water is 1000
kg/m3
. What is the density of the wood?
720
kg/m3
780
kg/m3
850
kg/m3
900
kg/m3

Answer 1

Answer on Question #51929-Physics-Field Theory

The specific heat of a substance at its boiling point or melting point

infinity

negative

between 0 and 1

Solution

The specific heat of a substance at its boiling point or melting point is infinity, because $\Delta T = 0$ but $\Delta Q \neq 0$ at these points in the formula for specific heat:

$$C = \frac {\Delta Q}{m \Delta T} \sim \frac {1}{\Delta T} \rightarrow \infty .$$

Answer: infinity.

10 Mass of gas is $m = 300\,g = 0.3\,kg$ and its specific heat at constant volume is $750\, \mathrm{J/kg\,K}$. If gas is heated through $75{}^{\circ}\mathrm{C}$ at constant pressure of $105\, \mathrm{N/\{m^2\}}$, it expands by volume $\Delta V = 0.08 \cdot 10^{6}\, \mathrm{cm}^{3} = 0.08\, \mathrm{m}^{3}$. Find CP/CV.

1.4

1.374

1.474

1.5

Solution

$$m C _ {P} \Delta t = m C _ {V} \Delta t + p \Delta V.$$

Thus

$$\frac {C _ {P}}{C _ {V}} = 1 + \frac {p \Delta V}{m C _ {V} \Delta t} = 1 + \frac {1 0 ^ {5} \cdot 0 . 0 8}{0 . 3 \cdot 7 5 0 \cdot 7 5} = 1. 4 7 4.$$

Answer: 1.474.

11 A solid ball with a mass $m = 0.53\, \mathrm{kg}$ floats in a tank of water. The ball is made of material with a density of $400\, \mathrm{kg/m^3}$. The density of water is $1000\, \mathrm{kg/m^3}$. What fraction of the volume of the ball is below the waterline?

0.1

0.2

0.3

0.4

Solution

The fraction of the volume of a floating object that is below the fluid surface is equal to the ratio of the density of the object to that of the fluid.

So,

$$\frac {V _ {\text {below}}}{V} = \frac {400 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}}}{1000 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}}} = 0.4.$$

Answer: 0.4.

12 A slab of wood with mass $m = 1.7$ kg floats $78\%$ submerged. The density of water is $1000 \, \text{kg/m}^3$. What is the density of the wood?

720 kg/m³

780 kg/m³

850 kg/m³

900 kg/m³

Solution

The fraction of the volume of a floating object that is below the fluid surface is equal to the ratio of the density of the object to that of the fluid.

So,

$$\frac {V _ {\text {below}}}{V} = 0.78 = \frac {\rho_ {\text {wood}}}{\rho_ {\text {water}}}.$$

Thus,

$$\rho_ {\text {wood}} = 0.78 \cdot 1000 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}} = 780 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}}.$$

Answer: $780 \frac{\mathrm{kg}}{\mathrm{m}^3}$.

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