Answer: 100% (double)
Explanation:
The question tells us two important things:
- Mass remains constant
- Volume remains constant
(We can think in a gas enclosed in a closed bottle, which is heated, for instance)
In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:
- State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1
- State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2
But we know that V1=V2 and that n1=n2, som dividing both sides, we get:
P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.
This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.
Answer:
Answer is B A cooler full of ice chilling a soda sitting on top of it
Explanation:
A cooler full of ice chilling a soda sitting on top of it
Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
Answer:
D
Explanation:
in series circuit the resistance is divided Total resistance is equal to the sum of resistances
Answer:
winter viscosity grades
Explanation:
The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.
