Answer:
the heat loss from this insulated wire is less
Explanation:
Given data in question
diameter of cable (d) = 20 mm
( K ) = 1 W/m-k
heat transfer coefficient (h) = 50 W/m²-K
To find out
the heat loss from this insulated wire
solution
we will find out thickness of wire
heat loss is depend on wire thickness also
we have given dia 20 mm
so radius will be d/2 = 20/ 2 = 10 mm
Now we find the critical thickness i.e.
critical thickness = K / heat transfer coefficient
critical thickness = 1 / 50 = 0.02 m i.e. 20 mm
now we can see that critical thickness is greater than radius 10 mm
so our rate of heat loss will be decreasing
so we can say our correct option is (a) less
Answer:
Thickness = 0.265 mm
Explanation:
Further explanation is given in the attached document.
Answer: Basically it's a substance with no shape and it's one of the different states of water.
Explanation:
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
So we equate the COP of our heater with COP of Carnot heater
Rearrange the equation
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C