question and solution for test 1

Q2

(a) A pill consists of thicken oil layer 10 cm static on water at the height of 30 cm from the bottom part of pill. Density of oil is 600 kgm^-3. (ρwater = 1000kgm^-3, P atm = 1.013 x 10^³ Pa). calculate:

i) Pressure gauge at the interface of oil-water (point A).

P = ρgh

= (1000)(9.8)(0.1)

= 0.98 x 10³ Pa.

ii) Absolute pressure at the bottom area of pill.

P = Patm + ρgh

= (1.013 x 10³) + [(1000)(9.8)(0.3)]

= 3953 Pa.

(b) A pressure rocket tank (figure 2) consists of 0.25 m³ kerosene weighted 200 kg. The pressure of upper part of kerosene is, Pa = 2.01 x 10⁵ Pa. the force, Fb that is stressed toward the bottom of tank is 20kN with area of 0.08m². calculate:

i) Density, ρ of kerosene.

P = m/v

= 200 / (0.8)(0.25)

= 1 x 10⁴ kg/m².

ii) Pressure at the bottom of the tank, Pb.

P = F/A

= 20k / 0.08

= 250 x 10³ Nm^-2.

iii) Depth of kerosene by using Pb = Pa + ρgh

Pb = Pa + ρgh

h = Pb – Pa / ρg

= 250 x 10³ – 201 x 10³ / (1 x 10⁴)(9.8)

= 0.5 m.

Q3

(a) an evening news broadcast in Kuala Lumpur, the meteorologist indicated that the

Day’s higher temperature was 38 ͦC. what is the corresponding value of this temperature on

i) Fahrenheit scale.

TF = (1.8)Tc + 32

= (1.8)(38) + 32

= 100.4 ͦF.

ii) Kelvin scale.

TF = Tc + 273.15

= 38 + 273.15

= 311.15 ͦK.

(b) a physics student wants to cool 0.2 kg of green tea (drinking water). The initial

Temperature of the green tea is 27 ͦC. by adding an amount of ice, m ice of -25ͦ C into the water,the final temperature of the mixture is 0 ͦ C with all the ice melted and the heat capacity of the container is neglected. The latent heat of fusion for ice is Lf = 333 x 10³ J.kg^-1 and specific heat, c for ice and water are 2100 J.kg^-1. ͦ C respectively, calculate

i) the heat change, Q of the green tea.

Q = mwater . C . ∆T

= (0.2)(4200)(27.0)

= 226803.

ii) The heat change, Q of ice in the order to warm it from temperature of

-25 ͦ C to 0 ͦ C in the form of unknown ice mass, m ice.

Q = mice . C . ∆T

mice = Q / C . ∆T

= 22680 / (2100)(0.25)

= 0.432 kg.

iii) the heat, Q that required to melt this mass of ice (in the form unknown ice mass, m ice ).

Q = mice (Lf)

= 0.432 (333 x 10³)

= 143.856 x 10³

= 143.86 kJ.

iv) the mass of ice , m ice when it reach thermal equilibrium. (assume there is

no heat loss).

Q = mice . C . ∆T

Mice = 143.86kJ / (2100)(0 + 25)

= 2.74 kg.