TEMPERATURE AND HEAT

4th WEEK


1) Hydrogen may be liquefied at-253˚C under the pressure of 20 atm. What is temperature on the Fahrenheit scale?
tf = 1.8tc + 32
=1.8(-253)+32
= - 391˚


2) A rod 3m long is found to have expanded 0.091cm in length for a temperature rise of 60˚C. what is α for the material of the rod?

α = 1/L∘ (∆L)/(∆T)
= (0.091x 10⁻²m)/((3m)(60k))
=5.1 x 10⁻⁶K⁻ ¹




3) Starting at 20˚C, how much heat is required to heat 0.3kg of aluminum to its melting point and then to convert it all liquid?

 Q = mc ∆t + mL. the specific heat of aluminum c is 0.22 kcal/kg. ˚C, heat of fusion L is 76.8 kcal/kg, and the melting point is 660˚C

Q = 0.3 (0.22) (660˚-20˚) + 0.3 (76.8)
= 42.24 + 23.04
=65.3kcal


4) A certain 6-g bullet melts at 300˚C and has specific heat capacity of 0.20cal/g. ˚C and heat of fusion of 15cal/g. how much heat to melt the bullet if it is originally at 0˚C?

 The bullet must first be heated to 300˚C and then melted.
 Heat needed = (6g) (0.20cal/g.˚C) (300-0)˚C + (6g) (15cal/g)
= 450cal, or 1880J


5) Refer to prob (4) .what is the slowest speed at wich the bullet can travel if it is to just melt when suddenly stopped?
 1880J is to be supplied as
K= (mv²)/2
1880J =(1/2)(0.006kg)v²min
So that
Vmin= 790m/s.

TUTORIAL

Hydrostatic

Tutorial

1. Find the pressure due to the fluid at a depth of 76cm in still

Water (pw=1.00g/cm³)

P=pgh

=(0.001kg/m³)(9.8m/s²)(0.76m)

=7.448kpa = 7.5 kPa

(b)mecury.(p=13.6 g/cm³)

P=pgh

=(0.0136kg/m³)(9.8m/s²)(0.76m)

=1.01×10⁵N/m²=101.3×10³ N/m²

2. A weight piston confines afluid density p in a closed container. The combined weight of piston and weight is 200N, and the cross-sentional area of piston is A=8.0cm².Find the total pressure at point B if the mercury and h=25cm.What would an ordinary pressure gouge read at B?

P = F/A + Pgh (2000/8 x 10⁻⁴)+13600(9.8m/s²)(25×10⁻²)

=2500000+33354

= 2.8×10⁵Pa

P=2.8×10⁵Pa+Pm

=2.8×10⁵Pa+1.01×10³Nm

=3.8×10⁵Pa

Pm=(13.6g/cm³)(9.81)90.76)

=0. 136kg/m³)(9.8m/s²)(0.76m)

=1.01×10⁵Nm

3. A vertical test tube has 2.0 cm of oil (p=0.8g/cm³)flooting 8.0cm of water. What is the pressure at botton of the tube due to the fluid in it ?

P₁=Pgh

=800kg/m (9.81m/s²)(2×10⁻²)

=156.96

P₂=Pgh

=1000kg/m (9.81m/s²)(8×10⁻²)

=784.8

P₁+P₂ = 156.96+784.8

=0.94 kPa

4. The u-tube devuce connected to the tank. Wahat is the pressure in the tank if atmosphera pressure is 76cm of mercury? The density of mercury is 13.6g/cm³.

P= 76cm – 13.6g/cm³(9.81ms²)(5×10⁻²)

= 76cm – 13600kg/cm³(9.81ms²)(5×10⁻²)

= 101.325 kPa -66708

=94.65×10³

=95 kPa

5. The mass of a block of aluminiumis 25.0g. (a) what is its volume?(b0what will be the tension in a string that suspends the block when the block is totally submerged in water?the density of aluminium is 2700kg/m²?

V=m/p ft=vg(e-pal)

=0.025/2700 =(9.26×10⁻⁶×9.81)(1000-2700)

= 9.26 =0.154N

6. A solid aluminium cylinder with p= 2700kg/m³ has a measured of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB=V₁P₁а

PT = FB / V1a

FB+Fr=mg

FB=mg-Fr

=(0.067×9.81)-(0.045×9.81)

=0.65727-0.4415

=0.21582

V1 = m/p = 0.067 / 2700

=2.481×10⁻⁵

P₁=0.21582

=(2.481×10⁻⁵)(9.81m/s²)

=8.9×10²kg/m³

Elasticity

Tutorial

1.An iron rod 4.00m long and 0.500cm²in cross section stretches 1.00mm when amass of 225 Kg is hung from its lower end.compute young’s modulus for the iron.

σ = F/A = (225 kg) (9.8m/s²) / (0.500 X 10⁻⁴ m²)

=4.41×10⁷ Pa

ε = ∆s/S∘

= 1.00 x 10 ⁻³m / 4.00 m

=2.5×10⁻⁴

Y= stress/strain

= σ/ε

=4.41x10⁷ / 2.5x10⁻⁴

=1.764×10¹¹

=176 Gpa

2. A load of 50 kg is applied to the lower end of steel rod 80 cm long and 0.60 cm in diameter, How much will the rod strectch?

m=50 kg×9.81 j²=3×10⁻³m ℓ∘ = 80cm

d=0.6cm÷2 A=πj² = 80×10⁻²

=0.3 =πj²(3×10⁻³)² = 0.8

=2.83×10⁻⁵m y=190kPa

Aℓo=Fℓo / Y∆

= (4.90 x 0.8)/(190x 10⁹)(2.83x 10⁵)

=392.4 / 5.377 x 10⁸

=72.97×10⁻⁷

=73μm

3. A platform is suspended by four wires at its corner. The wires are 3.0m long and lave diameter of 2.0mm. Young’s modulus for material is 180GPa.How far will the platform drop (due to elongation of the wires)if a50kg load is placed at the center of the platform?

Y=180GPa

=180×10⁹ Pa

D = 2mm÷2

=1×10×⁻³

Y = Fℓo / YA

ℓo=3m A=πj²

F=50×9.81 = (3.14×10⁻⁵)m

=490.5N

Aℓo=490.5N (3m²) / (180)(3.14 x 10 ˉ⁶)

=1471.5Nm/5.652x 10⁻⁴

=2.6×10⁶m

=2.6/4

=0.65mm

4. Two parallel and opposite forces,each 4000N,aree aplied tangentialy to the upper and lower faces of a cubical metal block 25cm on a side.Find the angel of shear and the displament of the upper surface relative to the lower surface.The shear modulus for the metl is 80GPa.

tanθ=2 x 10ˉ⁷ / 25 x 10⁻²

= 8.0×10⁻⁷ rad

Displament=4000 / (2.5 x 10ˉ²)

80GPa = 64000/∑

∑=64000/80G

=8×10⁻⁷(25×10⁻²)

=2×10⁻⁷

5.The bolk moduls of water is 2.1 Gpa. Compute the volume comtraction of 100ml of water when subjected to a pressure of 1.5 Mpa.

B= 2.1GPa

Vo= 100mL / 1000

=0.1 mL

∆p = 1.5Mpa

∆v = (∆p)(Vo) / B

=(1.5 Mpa)(0.1mL) / 2.1 GPa)

=150000 / 2.1GPa

= -71.43×10³

= -0.0714mL

6. The compressibility of water is 5.0×10⁻¹⁰m²/N,find the decrease in volume of 100ml of water subjected to a pressure of 15mPa.

Bulk strains = ∆Vo / Vo

AVo=B×VoA

=15×10⁶Pa(5.0×10⁻¹Nm⁻²)

=7.5×10⁻³

=0.75mL

ELASTICITY & HYDROSTATIC

QUESTION 3th WEEK

QUESTION 1 ;

1. The height of the mercury column in barometer is 760 mm. Find the pressure of the atmosphere in pascals.

ANSWER 1 ;

1. p = pgh
= ( 13.6 X 10 ³ kg/ m³ ) ( 9.80 m/s ² )( 0.760 m)
= 1.013 X 10 ⁵ Pa.

QUESTION 2 ;

2. A metal ball weighs 0.096N. When suspended in water it has an apparent weight of 0.071N. Find the density of the metal.


ANSWER 2 ;

2. The desired density is given by p = m/V. But since the volume V of the ball is also the volume of displaced water, the buoyant force is given by B = p water g V. Thus,
p = (mg)p water / B
= (0.096 N)(1 X 10³ kg/m³ ) / ( 0.096-0.071) N
= 3840 kg/m³ .

QUESTION 3 ;

3. Air has a density of 1.29kg/m ³ under standard conditions. What is the mass of air in a room with dimensions
10m x 8m x 3m?


ANSWER 3 ;


3. m = p V
V = (10 m)(8 m)(3 m) = 240 m ³
m = (1.29 kg/m ³ )(240 m ³ )
= 310 kg.

QUESTION 4 ;

4. Approximately how large a force is required to stretch a 2.0 cm diameter steel rod by 0.01 percent?
Y= 195000 M Pa


ANSWER 4 ;

4. F = AY(∆ L/L )
= ∏(0.01)² (195 X 10⁹ ) (10 ⁻⁴)
= 6100 N


QUESTION 5 ;

5. A copper wire 2.0 m long and 2 mm in diameter is stretched 1 mm. What tension is needed? Young's modulus for copper is 117600 M Pa


ANSWER 5 ;

5. F/A = Y ( ∆ L/ L)
F = Y ( ∆ L/ L) A
= (117.6 X 10⁹ ) 0.001 / 2 ( ∏ ( 0.001 )² )
= 184.7 N.