Consider the Conservation of Energy Equation: V² dE aq aw dt Ət Ət dmin + dt hin + [kJ-s] [kg-m²/s ³] in 2 + g Zin dmout dt where E is energy in [J], Q is heat in [J], W is work in [J], mis mass in [kg], h is enthalpy in [J/kg], Vis velocity in [m/s], g is gravitational acceleration is [m/s2] and z is height in [m]. A Watt is a Joule per second. Taking the derivative of a variable with respect time time is simply dividing the base variable unit by time. Determine, IN BASE UNITS ([kg], [m], [s]) the following: The units for enthalpy flux, i.e. the third term on the right-hand-side of the equation hout + [kg-m²/s²] Vont + 9%-out) 2 [kJ/s] [kJ-m²/s ³]

On Dec 5, 2022, scientific history was made at the Lawrence Livermore National Laboratory (LLNL) in Livermore, California when nuclear fusion was achieved when the 192 lasers deposited about 2 Megajoules (MJ) of energy into a frozen pea sized deuterium-tritium pellet and ignited the pellet through nuclear fusion to release 3 MJ of energy. The metric prefix mega means million (10^6). a) How much mass (kg) would be required to release 1 MJ of energy?

A hot tub with a surface area of 28 ft2 is filled with water to a depth of 29 in . Hint: volume is calculated as area × height (A × h). A) What is the volume of water in the tub, in liters? Express your answer to two significant figures and include the appropriate units. B) How many kilojoules are needed to heat the water from 59∘F to 103 ∘F? Express your answer to two significant figures and include the appropriate units. C) If the hot-tub heater provides 5900 kJ/min, how long, in hours, will it take to heat the water in the hot tub from 59∘F to 103∘F? Express your answer to two significant figures and include the appropriate units.

(3) The speed of light in vacuum, c, is 3 X 105 km/s and 1 kilometer = 1000 meters. How many Joules of energy are released if 2 kg of matter are converted into energy at 0.1% efficiency? First convert c to units of meters/second: c = 3 x 105 km/s x 1000 m/km = 3 x 108 m/s. Then express efficiency (eff) in decimal terms: eff = 0.001. In this type of problem, the mass/ energy equivalence equation is written as: E = (eff) mc2 Joules,