heat released Q = 749 joules
heat of fusion of silver L = 109 J/g
Here phase of silver is changing from liquid to solid
so temperature will remain same
all heat will be released due to its phase change
and in this case we use Q=mL
where m is the mass of silver in gram
Q= mL
749 = m * 109
m = 749/109
m = 6.87 gram
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer: D) Earthquake
<em>I hope this helps, and Happy Holidays! :)</em>
Answer:
a) 
b) 
Explanation:
a)
Given:
amount of heat transfer occurred,
initial temperature of car, 
final temperature of car, 
We know that the change in entropy is given by:
(heat is transferred into the system of car)
b)
amount of heat transfer form the system of house, 
initial temperature of house, 
final temperature of house, 
Answer:
13807.2 J/g°C
Explanation:
I just took the test and got it correct
