Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :
Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,
So, 12540 J of energy is used to heat the water.
Answer:
a) [volts] = [N m / C],
b) The lines or surface that has the same potential are called equipotential
c) the equipotential lines must also be perpendicular to the electric field lines
Explanation:
a) find the units of the volt
the electric potential energy is
V = k q / r
V = [N m² / C²] C / m
V = [N m / C]
The electric potential is defined as
V = E .s
V = [N / C] [m]
V = [N m / C] = [volt]
we see that in the two expressions the same result is obtained therefore the volt is
[volts] = [N m / C]
b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.
c) The electric potential is defined as the gradient of the electric field
v =
therefore the equipotential lines must also be perpendicular to the electric field lines
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
That is why the answer will be
,