Answer:
B. 180 million joules
Explanation:
Apply the formula for heat transfer given as;
Q=m*c*Δt where
Q = electrical energy consumed by the heater in joules
m= mass of air in the chamber in kg
c= specific heat of air in joules per kg degrees Celsius
Δt= change in temperatures in degrees Celsius
Given in the question;
m= 1200 kg
c= 1000 J/°C /kg
Δt = 180°-30°= 150° C
Substitute values in the equation to get Q as;
Q=m*c*Δt
Q= 1200 * 1000* 150
Q= 180000000 joules
Q = 180 million joules
<u>The correct answer option is B : 180 million joules.</u>
Answer:
The work required is W = 20.2 BTU per lbm
Explanation:
The value of entropy & enthalpy at initial conditions are
= 103.1
S = 0.225
Final enthalpy
= 123.3
Therefore work done
W = -
W = 103.1 - 123.3
W = - 20.2 BTU per lbm
Therefore the work required is W = 20.2 BTU per lbm
Answer:
Technician B only is correct
Explanation:
Voltage drop testing is a method used to find the amount of electrical resistance available in an high amperage circuit that involves connecting the leads of the meter in parallel to the circuits being tested such that disassembly is not required
In voltage drop test, the red voltmeter lead and black voltmeter lead are placed at two points on the same side of the circuit connection such that the leads are in between two positive connection or two negative connection (rather than connecting the red to the positive and the black to the negative sides of the circuit) and digital voltmeter is used for the voltage drop measurement across the lead while the connection is under load.
Therefore, Technician B only is correct
