Temperature and heat

Temperature and heat

Question 1

The Eiffel Tower, constructed in 1889 by Alexandre Eiffel, is an impressive latticework structure made of iron. If the tower is 301m high on a 22⁰C day, how much does height decrease when the temperature cools to 0.0⁰C?

Solution

∆L = αL0∆T

= (12 × 10⁻⁶ k⁻¹)(301 m)(-22K)

= -7.9cm

Question 2

The heat capacity of 1.00kg of water is 4186J/K.what is the temperature change of the water if(a)505 J of heat is added to the system, or (b)1010 J of heat is removed?

Solution

∆T = Q/C

= 505J/(4186J/K)

= 0.121K

∆T = Q/C

= -1010J/(4186J/K)

= 0.121K

Question 3

Calculate the radiated power from a sphere with a radius of 5.00 cm at the temperature 355 K. Assume the emissivity is unity.

Solution

P = eσ AT⁴

= (1)[5.67 × 10⁻⁸W/(m²·K⁴)]4π(0.0500m)²(355k)⁴

= 28.3W

Question 4

On New Year’s Day several human “polar bears” prepare for their annual dip into the icy waters of Narragansett Bay. One of these hardy soyls has a surface area of 1.15m² and a surface temperature of 303 K( -30⁰C). Find the net radiated power from this person (a) in a dressing room , where the temperatura is 293K(-20⁰C), and (b) outside, where the temperature is 273K (-0⁰C). Assume an emissivity of 0.900 for the person’s skin.

Solution

Pnet = eσ(T⁴-Ts⁴)

= (0.900)[5.67×10⁻⁸W/(m²·K⁴)](1.15m²)×[(303K)⁴ - (293K)⁴]

= 62.1W

Pnet = eσ(T⁴-Ts⁴)

= (0.900)[5.67×10⁻⁸W/(m²·K⁴)](1.15m²)×[(303K)⁴ - (273K)⁴]

= 169W

Question 5

(a) Suppose the volume V in Fig,15-7 is 30.0ml, while taht of V is 200 ml. What fraction, x, of the mass of the liquid has vaporized(at pressure P) when the volume is V = 100mL?

(b) If the heat of vaporization of this fictitious substance is 600kJ/kg and the mass is 80 g, how much heat must be added to the 30.0mL of liquid to reach the volume 100mL?

Solution

V = 100mL

= (1-x)(30.0mL)+x(200mL)

= 0.41

Amount of liquid is (0.41)(80 g) = 32.8g

= heat = (600kJ/kg)(0.0328 kg)=19.7kJ

Hydrostatic

Question

1. Atmospheric pressure is about 100kPa. How large a force does the air in a room exert on the inside of a window pane that is 40cm x 80cm?
The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and I given by
F = pA
= (100 kN/m²)(0.40 x 0.80 m²)
= 32 kN.


2. Find the ratio of a systolic blood pressure of 120 (in mmHg) to atmospheric pressure. Standard atmospheric pressure is 1.01 x 10⁵ Pa.
Pressure due to 0.120m of mercury = ρgh = (136000)(9.8)(0.120)
= 0. 16x10⁵ Pa
Ratio = 0.16 / 1.01
= 0.16


3. A hydraulic lift is to be used to lift a truck weighting 5000lb. if the diameter of the large piston of the lift is 1 ft, what gauge pressure in Ib/in² must be applied to the oil?
4. The gauge pressure in the oil acts, by Pascal’s Principle, on the bottom of the large piston to produce the force that lifts the load:
∆p = F/A
= 5000 / πr²
= 5000 / π(0.5²)
= 6370 Ib /ft²
= 6370 / 144
= 44.2 Ib/in²


4.Find the pressure exrted on the skin of a balloon if younpreess with a force 2.1N using (a) your finger or(b) a needle. Assume the area of yuor fingerip is 1.0 ×10⁻⁴m²,and the area of the needle tip is 2.5×10⁻⁷m².(c) Find the minimum force necessary to pop the balloon with the needle, given thatbthe balloon pops with a pressure of 3.0×10⁵N/m².

Solution
(a) P=F/A=2.1N/1.0×10⁻⁴m² = 2.1×10⁴N/m²
(b) P=F/A=2.1N/2.5×10⁻⁷m² = 8.4×10⁶N/m²
(c) F=PA
F=(3.0×10⁵N/m²)(2.5×10⁻⁷m²)=0.075N


5.A cubical box 20.00 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105.0 kPa;at the bottom the pressure is 106.8 kPa. What is the fluid?

Solution
ρ = P₂-P₁/gh
ρ =1.068×10⁵Pa-1.050×10⁵Pa/(9.81m/s²)(0.2000 m)
=920 kg/m³


6.A steel beam used in the construction of a bridge is 10.2 m long with a cross- sectional area of 0.12m². it is mounted between two concrete aboutments with no room for expansion. When the temprature rises 10⁰c, such a beam will expand in lenght by 1.2mm if it is free to do this expansion from happening?. Young ‘s modulus for steel is 2.0×10¹¹ N/m².

Solution
F = Y(∆L/Lo)A
Young modulus = 2.0×10¹¹ N/m²
∆L = 1.2mm – convert to (m²) = 1.2×10⁻³m
Lo = 10.2 m
A = 0.12 m²

F = (2.0×10¹¹ N/m²)(1.2×10⁻³m/10.2 m)(0.12m²)
=2.824×10⁶ N


7.A cube of jell-o 6cm on a side sits on your plate. You exert a hurizontal furce of 0.20N on top surface parallel to the surface and observe a side ways duspalacement of 5mm.what is the shear modulus of the jell-o?

Solution
Shear modulus, S=(F/A)/(∆x/h)
S = F/h∆n =0.20N/(6×10⁻²m)(5×10⁻³)
=0.20N/3×10⁻⁴
=0.667kNm⁻²



8.A 500N object is hung from the end of awire of cross-sectional area 0.010×10⁻⁴m² as in agure(a).the wire stretches from its origanal lenght of 200.00×10⁻²m to 250.00×10⁻²m.

(i) What is the stress on the wire?
Solution
Stress= F/A
=500N/0.010×10⁻⁴m³
=500×10⁶Nm⁻²

(ii) What is the strain on the wire?
Solution
Strain = ∆L/Lo
=200.00×10⁻²m - 250.00×10⁻²m.
=0.5×10⁻²m
=0.5×10⁻²m/200×10⁻²
=2.5×10⁻³m
(iii) Determine the Young’s modulusof the wire?
Solution
Young’s modulus, Y=(F/A)/ (∆L/Lo)
=500×10⁶Nm⁻²/2.5×10⁻³m
= 2×10¹¹N




9.When an object is submerged in the ocean to adepth of 3000m, the pressure increases by about 3030N/m² decrease in volume when ivnered to this depth? The bulk modulus of aluminum is 7×10¹⁰N/m²

solution

∆V = PV/B
∆V = (3030N/m² × 0.30m³) / 7×10¹⁰ N/m³
=909m³/7×10¹⁰
= 12.99× 10⁻⁹ m³



10.Find the exerted on the palm of your hand by atmospheric pressure.assume yuor palm measures 0.080 m by 0.10m.

Solution


F = PatA
= (1.01×10⁵Pa)(0.080m)(0.10m)
= 810 N

Elasticity



Question:

A telephone wire 120m long and 2.2mm in diameter is stretched by a force of 380N. What is the longitudinal stress? If the length after stretching is 120.10m, what is the longitudinal strain? Determine Young’s modulus for the wire.

Solution:

The cross-sectional area of the wire is……

A = πD² / 4
= π (2.2 x 10^-3 m)² / 4
= 3.8 x 10^-6 m²

Stress = F/A
= 380 N / (3.8 x 10^-6 m²)
= 100 x 10^6 N/m²
= 100 MPa

Strain = ∆l / l
= 0.01 m / 120 m
= 8.3 x 10^-4

Y = stress / strain
= 100 MPa / (8.3 x 10^-4)
= 120,000 Mpa


Question

The density of aluminum is 2.70g/cm³. What volume does 2.0kg occupy?

Solution:

Ρ = m/V or V = m/ρ
= 2000g / 2.70g/cm³
= 741 cm³





Question

The platform is suspended by four wires at its corners. The wires are 3m long and have a diameter of 2.0mm. Young’s modulus for the material of the wires is 180000Mpa. How far will the platform drop (due to elongation of the wires) if a 50kg load is placed at the center of the platform?

Solution:

• ∆L = (3m)(123N) / (3.14x10^-6m²)
= 65x10^-5m
= 0.65mm



Question
Determine the fractional change in volume as the pressure of the pressure of the atmosphere (0.1Mpa) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is 125000 Mpa.

Solution:

B = ∆V / V
= -∆p / B
= - (-0.1) / 125000
= 8x10^-7


Question

Compute the volume change of solid copper cube, 40mm on each edge, when subjected to a pressure of 20 Mpa. The bulk modulus for copper is 125000Mpa.

Solution:

∆V = -V∆p / B
= -(40mm)³(20Mpa) / 125000Mpa
= -10mm³



Question

To inspect a 14,500 Ncar,it is raised with a hydraulic lift.if the radiusof the small piston is 4.0cm,and the radius of the large piston is 17cm, find the force that must be exerted on the small piston to lift the car.


Solution


F₁ = F₂(A₁/A₂)
= (14,500 N)(π(0.040m)²/(π(0.17m)²)
= 800 N

ELASTICITY

Question:

A telephone wire 120m long and 2.2mm in diameter is stretched by a force of 380N. What is the longitudinal stress? If the length after stretching is 120.10m, what is the longitudinal strain? Determine Young’s modulus for the wire.

Solution:

The cross-sectional area of the wire is……

A = πD² / 4

= π (2.2 x 10^-3 m)² / 4

= 3.8 x 10^-6 m²

Stress = F/A

= 380 N / (3.8 x 10^-6 m²)

= 100 x 10^6 N/m²

= 100 MPa

Strain = ∆l / l

= 0.01 m / 120 m

= 8.3 x 10^-4

Y = stress / strain

= 100 MPa / (8.3 x 10^-4)

= 120,000 Mpa

Question

The density of aluminum is 2.70g/cm³. What volume does 2.0kg occupy?

Solution:

Ρ = m/V or V = m/ρ

= 2000g / 2.70g/cm³

= 741 cm³

Question

The platform is suspended by four wires at its corners. The wires are 3m long and have a diameter of 2.0mm. Young’s modulus for the material of the wires is 180000Mpa. How far will the platform drop (due to elongation of the wires) if a 50kg load is placed at the center of the platform?

Solution:

· ∆L = (3m)(123N) / (3.14x10^-6m²)

= 65x10^-5m

= 0.65mm

Question

Determine the fractional change in volume as the pressure of the pressure of the atmosphere (0.1Mpa) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is 125000 Mpa.

Solution:

B = ∆V / V

= -∆p / B

= - (-0.1) / 125000

= 8x10^-7

Question

Compute the volume change of solid copper cube, 40mm on each edge, when subjected to a pressure of 20 Mpa. The bulk modulus for copper is 125000Mpa.

Solution:

∆V = -V∆p / B

= -(40mm)³(20Mpa) / 125000Mpa

= -10mm³

Hydrostatic.

tutorial

1. Find the pressure due to the fluid at a depth of 76 cm in still
Water (ρw = 1.00g/cm^3)
Mercury (ρ = 13.6g/cm^3)
Solution :
P = ρgh
= (1000 kg/m^3) (9.8m/s^2) (0.76)
= 7.448kPa @ 7.5kPa
P = ρgh
= (13600 kg/m^3) (9.8m/s^2) (0.86)
= 1.01 x 10^5Nm^2 @ 101.3 x 10^3N/m^2

2. A weight piston confines a find density ρ in a closed continue. The combined weight of piston and weight iss 200N, the total cross-sectional area of the piston is A = 8.0cm^2. Find the total pressure at point B if the fluid is mercury and h = 25cm. What would and ordinary pressure gauge read at B?

Solution :
P = (F/A) + ρgh
= (2000/8 x 10^ -4) + (13 600) (9.81m/s^2) (25 x 10^ -2)
= 250 000 + 33 354
= 2.8 x 10^5 Pa
P m = (13.6g/cm^3) (9.81m/s^2) (76cm)
= (0.136kg/m) (9.81m/s^2) (0.76cm)
= 1.01 x 10^5N/m
P = (2.8 x 10^5Pa) + Pm
= 92.8 x 10^5Pa) + (1.01 x 10^3N/m)
= 3.8 x 10^5Pa

3. A vertical test tube has 2.0cm of oil (ρ = 0.8g/cm^2) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?

Solution :
P1 = ρgh
= (800kg/m) (9.81m/s^2) (2 x10^ -2)
= 156.96
P2 = (1000) (9.81m/s^2) (8 x 10^ -2)
= 784.8
P1+P2 = 156.96 + 784.8
= 0.94kPa

4. The U-tube device connected to the tank. What is the pressure in the tank if atmosphera ρ is 76cm of mercury? The density of mercury is 13.6kg/m^3?

Solution :
P = (76cm) – (13.6g/cm^3) (9.81m/s^2) (5 x 10^ -2)
= (76cm) – (13 600kg/cm^3) (9.81m/s^2) (5 x 10^ -2)
= 94.65 x 10^3
= 95kPa


5. The mass of a block of a aluminium is 25.0g.
What is its volume?
What will be tthe tension in a string that suspends the block when the block is totally submerged in water? The density of aluminium is 2700kg/m^3.
Solution :
V = m/p
= (0.025) / (2700)
= 9.26cm^3
ft = v,( l-Pal )
= (9.26 x 10^ -6) (9.81) (1000-2700)
= 0.154N


6. A solid aluminium cylinder with ρ=2700kg/m^3 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine?

Solution :
Pt = RB / VTg
FB+FR = mg
FB = mg – FR
= (0.067 x 9.81) – (0.045 x 9.81)
= 0.21582
Vr = m / p
= (0.067) / (2700)
= 2.48 x 10^ -5
Pt = 0.21582
= (2481 x 10^ -5) (9.81)
= 886.7
= 8.9 x 10^2 kg/m^3

Elasticity.

tutorial.

1. An iron rod 4.00m long an 5.00cm^2 in cross section stretches 1.00mm. When a mass of 22.5kg is hung from its lower end. Compute Young’s Modulus from the iron.

Solution :

σ= F/A
= (225kg) (9.8m/s^2) / (0.500 x 10^ -4)
= 4.41 x 10^7 Pa

ε = Δƪ / ƪ˳
= (1.00 x 10^ -3 m) / (4.00m)
= 2.5 x 10^ -4

Y = σ / ε
= (4.41 x 10^7 Pa) / (2.5 x 10^ -4)
= 1.764 x 10^11
= 176 GPa

2. A load of 50 kg is applied to the lower end of a steel rod 8cm long and 0.60cm in diameter. How much will the rod sterch Y = 190 GPa for stell?

Solution :

Formula : Y = F ƪ˳ / AΔ ƪ A ƪ = F ƪ˳ / Y. A

m = 50 kg x 9.81
d = 0.6 cm / 2
= 0.3
j^2 = 3 x 10^ -3m
A = п j^2
= п (3 x 10^ -3)^2
= 2.83 x 10^ -5

ƪ˳ = 80 cm
= 80 x 10^ -2
= 0.8
Y = 190 GPa
Aƪ˳ = F ƪ˳ / YA
= (490.5) (0.8) / (190 x10^9) (2.83 x 10^5)
= 392.4 / 5.377 x 10^8
= 72.97 x 10^ -7
= 73 x 10^ -6
= 73μm

3. A platform is suspended by four wire at its corner. The wire are 3.0m long and have a diameter at 2.0mm Young’s Modulus for the material is 180GPa. How for will the platform drop (due to elongation of the wire) if a 50kg load is placed at the center of the platform?

Solution :

ƪ˳ = 3m
F = 50 x 9.81
= 490.5N
A = пj^2
= (3.14 x 10^ -6)m
A ƪ˳ = 490.5Nm / (5.652 x 10^ -4)
= 2.6 x 10^6
= 0.65m

4. Two parallel and opposite force. Each 400N are applied tangentially to the upper and lower forces of a cubical matel block 25cm on a side. Fine the angle of shear and the displacement of the upper surface relative to the surface. The shear modulus for the metal is 80GPa.

Solution :
F = 400N
h = 25cm
A = 0.0625m^2
S = 80 x 10^4 p
tanθ = (2 x 10^ -7) / (25 x 106 -2)
= 8.0 x 10^ -7 rad
x = FAh / SA
x = (4000) (0.25 x 10^ -3) / (8.0 x 10^6Pa) (0.0625)
= 2 x 10^ -7m

5. The bulk modulus of water is 2.1GPa. Compute the volume contraction of 100ml of water when subjected to a pressure of 1.5mPa?

Solution :
Av˳ = - Pv˳ / B
B = - Pv˳ / Δ v˳
B = 2.1GPa
P = 1.5MPa
= 1.5 x 10^6Pa
v˳ = (100ml) / (1000)
Δ v˳ = - Pv˳ / B
= (1.5 x 10^6) (0.1) / (2.1 x 10^9)Pa
= (-7.14 x 10^ -5mm) x (1000)
= -0.0714ml

6. The compressibility of water is 5.0 x 10^ -10m^2/N. Find the decrease in volume of water of 100ml when subjected to a pressure of15mPa?

Solution :

A v˳ = B x v˳
= (15 x 10^6) (5.0 x 10^ -10Nm^ -2)
= 7.5 x 10^ -3) x(100)
= 0.75ml