Answer:
a)
, b) 
Explanation:
a) The heat transfered to the egg is computed by the First Law of Thermodynamics:





b) The amount of entropy generation is determined by the Second Law of Thermodynamics:



Answer:
The mass percentage of calcium nitrate is 31.23%.
Explanation:
Let the the mass of calcium nitrate be x and mass of potassium chloride be y.
Total mass of mixture = 19.12 g
x + y = 19.12 g..(1)
Mass of solvent = 149 g = 0.149 kg
Freezing point of the solution,
= -5.77 °C
Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)
The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:
i = 3
i' = 2
Freezing point of water = T = 0°C




On solving we get:
....(2)
Solving equation (1)(2) for x and y:
x =5.973 g
y = 13.147 g
Mass percent of
in the mixture:

The mass percentage of calcium nitrate is 31.23%.
Answer:
Required energy = 4758 J
Explanation:
Specific heat capacity of a material is the amount of energy required to raise the temperature of one kilogram (kg) of that material through one degree Celsius (°C).
Given data :
Specific heat capacity = c = 2440 J/kg.°C
Mass = m = 150 g = 0.15 kg
Initial temperature = 22°C
Final temperature = 35°C
Change in Temperature = ΔT = 13°C
Energy = E = ?
Using the following formula and substituting the values, we get:
E = m × c × ΔT
E = 0.15 × 2440 × 13
E = 4758 J
Answer:
(a) The density of the object is 316/343 × the density of the oil
(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume
Explanation:
(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

The volume of the object is therefore;

Where 6 cm is above the oil level we have;
above the oil level
Therefore, volume of the oil displaced =
cm³ = 216.11 cm³
The density of the object is thus;

The density of the object = 316/343 × the density of the oil.
(b) The volume of the oil = 2 × Volume of the object = 
The fraction of the volume displaced, x, after immersing the object is given as follows;

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil
A) Vt = Vo - gt
= 0 - 9.8 . 2
= - 19.6 m/sec (acting downward)
b) ∆y = Vt^2 - Vo^2 / 2g
= (-19.6)^2 - 0 / 19.6 = 19.6 meters