Answer:
The temperature attains equilibrium with the surroundings.
Explanation:
When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.
Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.
Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.
Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 ×
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 ×
temperature = 25°C
surface heat transfer coefficient = 6 6 W
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4 × r1 × r2 × K
R1 = ( 5.2 - 5 ) / 4 × 5 × 5.2 × 16 ×
R1 = 3.825 ×
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4 r2²
area = 4 (5.2 × )² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 × + 4.90499
heat loss = 2.037152 W
Answer:
The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is an idication that your vehicle may be developing a cooling system problem.
Explanation:
Answer:
If your spent lead-acid batteries will be reclaimed through regeneration, you are exempt from all of the hazardous waste regulations except for Part 261 and §262.11 (both of which are waste identification requirements). ... The Universal Waste rules in Part 273 apply to all types of batteries that would be hazardous waste
Explanation:
Answer:
Rf=470 ohms R1=10ohms, Schmatic is attached
Explanation:
For non-inverting amplifiers
Gain= 1+(Rf/R1)
480=1+(Rf/R1)
470=Rf/R1
if Rf=470Ω and R1=10Ω
then,
47=47