Answer:
(a) ΔU=747J
(b) γ=1.3
Explanation:
For (a) change in internal energy
According to first law of thermodynamics the change in internal energy is given as
ΔU=Q-W
Substitute the given values
ΔU=970J-223J
ΔU=747J
For(b) γ for the gas.
We can calculate γ by ratio of heat capacities of the gas
γ=Cp/Cv
Where Cp is the molar heat capacity at constant pressure
Cv is the molar heat capacity at constant volume
To calculate γ we first need to find Cp and Cv
So
For Cp
As we know
Q=nCpΔT
Cp=(Q/nΔT)

From relation of Cv and Cp we know that
Cp=Cv+R
Where R is gas constant equals to 8.314J/mol.K
So

So
γ=Cp/Cv
γ=[(37J/mol.K) / (28.687J/mol.K)]
γ=1.3
Answer:
(a) = -0.16%
(b) = smaller
Explanation:
given
power = 460 W
potential difference = 120 V
(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?
we know
.....................(i)
we need to find change in power
..............(ii)
from equations we get



(b)
if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power
and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller
Answer:
See the answers below.
Explanation:
The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.
![Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]](https://tex.z-dn.net/?f=Cost%3D0.350%5BkW%5D%2A12%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%2A30%5Bdays%5D%2A4.5%5B%5Cfrac%7BRs%7D%7BkW%2Ahr%7D%20%5D%3D567%5BRs%5D)
The fuse can be calculated by knowing the amperage.

where:
P = power = 350 [W]
V = voltage = 240 [V]
I = amperage [amp]
Now clearing I from the equation above:
![I=P/V\\I=350/240\\I=1.458[amp]](https://tex.z-dn.net/?f=I%3DP%2FV%5C%5CI%3D350%2F240%5C%5CI%3D1.458%5Bamp%5D)
The fuse should be larger than the current of the circuit, i.e. about 2 [amp]