how to solve the following

Question

7.2. a piece of cast iron is 7cm X 13cm X 9cm and is heated from a temperature at 4 degrees Celsius, to 50 degrees Celsius. The coefficient of linear expansion of the cast iron is 14X10^6 degrees Celsius. 7.3. 4 kg of air with a gas constant of 289 J/kg.K is compressed in a container. The volume is 0,134m^3 and the pressure is 600kPa. 7.4. the temperature of a steel rod with a diameter of 43mm and length of 0.131 m/s increased by 83 degrees Celsuis. The increase in volume is 8131mm^3. Calculate the percentage increase in volume.

Answer 1

(7.4)

Length of the steel rod = 0.131 m.

Diameter of the steel rod = 43 mm = 0.043 m

Volume of the steel rod = Cross sectiona area * length

Cross-sectinal area = πr²

A = π(0.0215)²

A = (22/7) (0.00046225)

A = 1.45 8 10^-3 m².

Cross sectional Area = 1.45 * 10^-3 m²

Voume = 1.45 8 10^-3 * 0.131

V = 1.9 * 10^-4 m³.

Volume = 1.9 * 10^-4 m³.

Increased is increased by 8131 mm³

Percentage volume = (8131 mm³ / (1.9 * 10^-4 m³)) * 100

% v = ((8.131 *10^-6 m³)/(1.9 * 10^-4 m³)) * 100

% v = 0.0427 * 100

% v = 4.27 %

Increase in the volume of the steel rod = 4.27%

Answer 2

(7.2)

The dimensions of the cast iron is 7cm * 13cm * 9cm.

Then length of the cast iron = 7cm.

The temperature rises from 4°C to 50°C.

Linear coefficient of expansion = 14 * 10^-6

We know that

where

= 7 cm

= Initial Temperature 4°C

= Final Temperature 50°C

L1 - 0.07 = (0.07) * (14*10^-6) * (50-4)

L1 - 0.07 = 4.508 * 10^-5

L1 = 0.07004508 m

L1 = 7.004508 cm.

Length of the cast iron at 50°C is 7.004508 cm.

Answer 3

(7.3)

The mass of the gas = 4Kg

Gas Constant R = 289 J/Kg.K.

Volume of the 0.134 m³.

Pressure of the gas = 600 kPa

Ideal gas Equation is PV = mRT.

where P = Pressure of the gas.

V = Volume of the gas.

m = mass of the gas in Kg.

R = Gas constant = 289 J/Kg.K.

T = Temperature in Kelvin.

T = PV / mR

T = (600 * 10^3 * 0.134) / ( 4 * 289)

T = 69.55 K.

Therefore temperature of the gas = 69.55 K.