Answer:
C) 43,2°C
Explanation:
<em>Sensible heat</em> is the amount of thermal energy that is required to change the temperature of an object, the equation for calculating the heat change is given by:
Q=msΔT
where:
- Q, heat that has been absorbed or realeased by the substance [J]
- m, mass of the substance [g]
- s, specific heat capacity [J/g°C]
- ΔT, changes in the substance temperature [°C]
To solve the problem, we clear ΔT of the equation and then replace our data:
Q=890 [J],
m=16,6 [g],
s=2,74 [J/g°C]
Q=msΔT.......................ΔT=Q/ms
Δ
°C
As:
ΔT=Tfinal-Tinitial
Tfinal=ΔT+Tinitial
Tfinal=21,7+21,5=43,2°C
The final temperature of the ethanol is 43,2°C.
Answer:
Go on amazon and look for HDMI -> (VGA/DVI/HDMI) what ever you need adapters.
An ideal voltage source provides no energy when it is loaded by an open circuit (i.e. an infinite impedance), but approaches infinite energy and current when the load resistance approaches zero (a short circuit). ... An ideal current source has an infinite output impedance in parallel with the source.
Answer:
The tube diameter is 2.71 mm.
Explanation:
Given:
Open glass tube is inserted into a pan of fresh water at 20°C.
Height of capillary raise is four times tube diameter.
h = 4d
Assumption:
Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.
Take surface tension of water at 20°C as
N/m.
Take density of water as 100 kg/m3.
Calculation:
Step1
Expression for height of capillary rise is gives as follows:
Step2
Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:
m.
Or
d=2.719 mm
Thus, the tube diameter is 2.719 mm.
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit