Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.
F=k•x
F=20•9.81=196.2N
x=.12m
196.2=k•.12
k=1635 N/m
Answer:
745.4K ~ 472.3 C
Explanation:
This is an Ideal Gas Law problem where we have to manipulate the equation a bit. Let's start with the basic:
PV = nRT will be used for both the initial and final, so we will rearrange this problem to state:
(V(initial))/(T(Initial)) = nR/P
Since we know that the pressure, number of moles of He, and ideal gas constant (R) remain the same from start to finish so we can write the problem as such:
(V(initial))/(T(Initial)) = nR/P = (V(final))/(T(final))
or
(V(initial))/(T(Initial)) = (V(final))/(T(final))
Now lets define some of these values:
T(initial) = 25degree (assuming degrees Celsius) ~ 298.15K
V(initial) = 2.0L
V(final) = 5.0L
T(final) = ?
Since we are solving for T(final) let's rearrange the problem once more to be solving for T(final):
T(final) = (V(final)T(Initial))/V(initial)
Now plug in your values:
T(final) = (5.0L*298.15K)/(2.0L) ~ 745.4K ~ 472.3degrees Celsius
And the other answer is jeje