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Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Answer:
G.P.E = 368.3
Explanation:
Given the following data;
Mass = 2.63kg
Height, h = 14.29m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the gravitational potential energy;
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;

Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.

G.P.E = 368.3
Note: the unit of gravitational potential energy is Joules.
Frequency of light = speed of light/wavelength = 3x10⁸/725x10⁻⁹ =4.13 x 10¹⁴ Hz