0.0179 ohms for copper.
0.0184 ohms for annealed copper
Ď = R (A/l) where
Ď = electrical resistivity
R = electrical resistance of a uniform specimen
A = cross sectional area
l = length
Solve for R by multiplying both sides by l/A
R = Ď(l/A)
The cross section of the wire is pi * 1^2 mm = 3.14159 square mm = 3.14159e-6 square meters.
The length is 3 meters. So l/A = 3/3.14159e-6 = 9.5493e5
Ď for copper is 1.68e-8 so 1.68e-8 * 9.5493e5 = 1.60e-2 ohms at 20 C
But copper has a temperature coefficient (α) of 0.00386 per degree C.
So the resistance value needs to be adjusted based upon how far from 20 C the temperature is.
50 - 20 = 30 C
So 0.00386 * 30 = 0.1158 meaning that the actual resistance at 50 C will be 11.58% higher.
So 1.1158 * 0.016 = 0.0179 ohms.
If you're using annealed copper, the values for Ď and the temperature coefficient change.
Ď = 1.72e-8
α = 0.00393
Doing the math, you get
1.72e-8 * 9.5493e5 * (1 + 30 * 0.00393) = 0.0184 ohms
Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
Answer:
a) 6.4 x 10^-12 cm^3
b) 17 x 10^-6 mm^2
Explanation
a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3
3
4
πr
3
V = \frac{4}{3}*3.142* 1.15^3
3
4
∗3.142∗1.15
3
= 6.3715 μm^3
1 μm^3 = 10^-12 cm^3
6.3715 μm^3 = 6.3715 x 10^-12 cm^3
==> 6.4 x 10^-12 cm^3
Answer:
Explanation:
The average kinetic energy of molecules in a pot increases when it is heated on a stove . The heat energy is spent to decrease the intermolecular attractive force . The result is increase in kinetic energy due to increase in the velocity of molecules
formula for kinetic energy of one mole of a monoatomic gas
= 3/2 RT
when temperature increases , kinetic energy increases .