3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.
1 answer:
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
(1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):
Then, you replace the values of all parameters in the equation (1):
The amount of water losses a heat of 514 cal
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Answer:
(a) 1.87 x 10⁻⁴ m/s
(b) 0.013V
Explanation:
(a) Drift speed, , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;
=
----------------(i)
Where;
q = amount of charge
n = free charge density
A = cross-sectional area of the wire
But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e
J = =
Equation (i) can then be written as follows;
=
=
=
---------------------(ii)
From the question;
E = 0.0520N/C
p = 1.72 x 10⁻⁸ Ωm
n = 8.5 x 10²⁸ electrons/m³
c = charge on electron = 1.9 x 10⁻¹⁹C
Substitute these values into equation (ii) as follows;
=
= 1.87 x 10⁻⁴ m/s
(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e
V = E x d [where d = 25.0cm = 0.25m]
V = 0.0520 x 0.25
V = 0.013V
Answer:
1.95m/s
Explanation:
Please view the attached file for the detailed solution.
The following were the conversion factors used in order to express all quatities in SI units:
