Power(p) = 2kW
Time(t)= 3 h
Work done(w) = ?
p = w / t
w = p × t
w = 2kW × 3 h
w = 6kWh
Converting 6kWh to Joules,
6kWh
6×(10^3)h ( k = 10^3 )
6×10^3×(60×60) ( 1 hour = 60 × 60 sec)
2.2×10^7 J in 2 significant figures
Answer:
P = 942.54 W
Explanation:
Given:
Temperature, T = 310 K
mass, m = 70 kg
Now from the stefan's law,
we have the formula
P = σAT⁴
where,
P = radiate energy
σ = Stefan's constant = 5.67 × 10⁻⁸ W/(m² K⁴)
A = Area of the body
Average surface area for a human body = 1.8 m²
T = Temperature
on substituting the values we get,
P = 5.67 × 10⁻⁸ × 1.8 × 310⁴
or
P = 942.54 W
Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h
The refrigerant during expansion undergoes a throttling process
Therefore,
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia
Using formula
Put the value into the formula
Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.
It's B because when you throw something it doesn't go up it slowly descends downward