One kg of air is heated in a closed rigid vessel

AI Thread Summary
The discussion revolves around calculating the heat transfer and change in internal energy for one kg of air heated from 27°C to 427°C in a closed rigid vessel. The relevant equations include Q = m * Cv * (T2 - T1) for heat transfer and ΔU = Q for internal energy change, given that no work is done due to constant volume. The user initially calculates heat transfer but questions the unit correctness and the relationship between R and Cv. Clarification is provided that since the vessel is rigid, the heat transferred equals the change in internal energy, simplifying the calculations. The conversation emphasizes the importance of understanding the properties of gases under constant volume conditions.
manal950
Messages
177
Reaction score
0

Homework Statement



One kg of air is heated in a closed rigid vessel such its temperature changes from 27 C to 427 C . Find the heat transfered
and change in internal energy .
Assume : -
R= 0.287 KJ/kg k
Cv = 0786 KJ/Kg k

Homework Equations



Q = mXCX(T2 - T1 )

The Attempt at a Solution



please can anyone explain to me how I can solve this question

and is R and Cv must be in same unit

my answer :

heat transfer = 1 X 0.786 X 10 (700-300 )

314.4X 10^3 KJ ( is the unit correct )

and about second one I don't have any idea for solve it
 
Physics news on Phys.org
ΔU=q + w(external)
w=work
as you posted Cv then volume is constnt so change in pressure should be taken.so by Gay Lussacs law P=kT
thereforeP1/P2=T1/T2
so you can find Work by volume*change in pressure
add Q to get work
 
I don't understand clearly can please help ?
 
Since the container is rigid, there is no volume change and no work is done. The heat transferred is equal to the change in internal energy.
 
thanks so much
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Change in temperature of air in a closed system
Replies
16
Views
2K
Adiabatic Expansion of Pressurized Air in a Piston-Cylinder Setup
Replies
8
Views
2K
Specific work done by an adiabatic & reversible compressor on air
Replies
6
Views
2K
How long does it take to heat a vessel?
Replies
41
Views
5K
Molecular Specific Heat of an Ideal Gas: Computations
Replies
8
Views
2K
Why this special case occurs in an exchange of heat
Replies
5
Views
2K
Modeling a kongming lantern (sky lantern)
Replies
6
Views
2K
Specific heat capacity of a metal at low temperature
Replies
8
Views
6K
Specific heat capacity of a solid material
Replies
9
Views
2K
Thermodynamics Change in Internal Energy?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top