Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.
D. because Land does reach higher temperatures then water and Land heats more rapidly
Answer:
α = 17.0 rad/s²
Explanation:
For this problem let's use the torque expression
τ = I α
Torque is the product of force by the distance that is half the length of the pencil
τ = F x
Let's use trigonometry
sin 10 = x / (l / 2)
x = l / 2 sin 10
τ = mg l /2 sin 10
The moment of inertia of pencil that we approximate as a rod with the axis of rotation at one end
I = 1/3 m L²
We replace
mg l / 2 sin 10 = 1/3 m l² α
α = 3/2 g/l sin 10
Let's reduce the magnitudes to the SI system
l = 15.0 cm = 0.150 m
m = 10.0 g = 0.0100 kg
Let's calculate
α = 3/2 9.8 / 0.150 sin 10
α = 17.0 rad/s²
Answer:
(a) 1.35 x 10^6 Joule
(b) 321.45 Kcal
Explanation:
Energy consumed in 1 hour = 270 kJ
So, Energy consumed in 5 hour = 270 x 5 = 1350 kJ
(a) Energy in joules = 1350 x 1000 J = 1.35 x 10^6 J
(b) 4.2 Joule = 1 calorie
So, 1.35 x 10^6 Joule = 1.35 x 10^6 / 4.2 = 0.32 x 10^6 Calorie
= 0.32 x 1000 Kcal = 321.45 Kcal