In the SI system of units, the absolute temperature is measured to be 303 K Its value in Fahrenheit is a-) 76 F b)- 86F c)-79 F
1 answer:
Answer:
The correct answer is option B i.e. 86 f
Explanation:
The freezing point of water is 32°f and boiling point is 212 °f. Thus difference in boiling and freezing point is exactly come out to be 180°. therefore fahrenheit scale is 1/180 interval of both scale ( freezing and boiling point)
given data:
absolute temperature is 303 K
We know by general formula of temperature in fahrenheit
°F = × (K-273) +32
°F = × (303-273) +32
°F = 86 F
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Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K =
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
...........1
put here all these value and we get t2
t2 = 424.8
final temperature is 424.8 K
so correct option is e
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ