Answer:
W = 3/2 n (T₁- T₂)
Explanation:
Let's use the first law of thermodynamics
ΔE = Q + W
in this case the cylinder is insulated, so there is no heat transfer
ΔE = W
internal energy can be related to the change in temperature
ΔE = 3/2 n K ΔT
we substitute
3/2 n (T₂-T₁) = W
as the work is on the gas it is negative
W = 3/2 n (T₁- T₂)
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E =
+ Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
Answer:
The maximum no. of electrons- 
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,

Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:

Maximum number of electrons, n is given by:

where
e = charge on an electron = 
Thus

Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2
F = 9.00 * 10^9 N.m^2/C^2 * 2.4*10^-8 C * 1.8*10^-6 C / [0.008m]^2 = 38.88 * 10^ -5 N
F = 39 * 10 -5N