Answer : The rate of heat transfer to the water is, 37.92 kJ/min
Explanation : Given,
Time = 10 min
Mass of water = 200 g
Latent heat of fusion of water = 334 J/g
Latent heat of vaporization of water = 2230 J/g
Now we have to calculate the rate of heat transfer to the water.
Now put all the given values in the above formula, we get:
Thus, the rate of heat transfer to the water is, 37.92 kJ/min
Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object, 
Initial temperature, 
Final temperature, 
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :
Let the mass of the object is 10 g or 0.01 kg
So,
Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.
Answer:
b) Document lessons learned.
Explanation:
First he should do documentation
then C
then D
then A
When the metals touch
together, half the charge of the charged metal flows to the other because the
electrons all repel each other. Therefore this also means that each metal ball
contains the same amount of electrons. Each ball has 5^10 electrons, this is
equivalent to a total charge of:
Q1 = Q2 = (1.602 * 10^-19
coulombs / electron) 5^10 electrons = Q
Q = 1.564 * 10^-12 C
Now using the Coulombs
law to find for the electric force:
F = k q1 q2 / r^2 = k (Q)^2
/ r^2
where k is a contant = 9
* 10^9 N m^2 / C^2
r = the distance of the
two metals = 0.2 m
So,
F = (9 * 10^9 N m^2 /
C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2
F = 5.51 * 10^-13 N
Since the two metals
repel therefore they are the one which exerts the force hence the magnitude
must be negative:
<span>F = - 5.51 * 10^-13 N</span>
B. 1520 is the difference between their weights.
