Q=mc\Delta\Deltat
Q= (0.25)(2100)(0--30°C)
Q = 6300J
(now find the Power)
P=Q/\Deltat
P=6300J/150s
P=43W
Now I can use the first formula I figured out:
Q=mL_{f}
=(0.25kg)(330000J/(kg·°C)
=82500J
Next step would be to find the time!
\Deltatime=Q/P
\Deltat=82500J/43W
=1918.65...
I've tried many different formulas and most don't make sense to me :/
For example this is one of the formulas I've gotten, but it just seems so wrong to be correct because once you put it into the other equation you get like 3 seconds which makes NO sense what so ever.
\Deltat=Q/P
150s =...
Homework Statement
The mass of the piece of ice is 0.25kg.
Heated by an electric heater (assume there is no loss of energy to the surroundings)
Started at -30°C at 0seconds
After 150s the ice cube is -10°C
Homework Equations
Q=mL_{f}
\Deltat=Q/P
The Attempt at a Solution...
Vaporization since you want to vaporize the mass.
So with that being said
Q=(8.0kg)(4200J(kg·°C)(100°C-25°C)+.25(8.0kg)(2300000)
Q=2520000 + 4600000
Q=7120000J
I loved the pun, I actually had a giggle with that one!
Right now I'll have to find the mass of the steam itself!
Q=
To find the additional energy would I use the equation that I initially posted (would I use latent fusion or Latent vaporiztion)?
BUT I don't know the mass of the...
Would it be:
Q=mc\Deltat
Q=(8.0kg)(4200J/(kg·°C))(100°C-25°C)
Q=2 520 000
Then since your finding a a quater of it would you then muliply by .25
2520000x.25
=630000J
It seems to easy since the question is worth 5 marks, it feels like I'm missing a step or something :/
Homework Statement
How much energy is required to change one quarter of 8.0kg of water at 25°C into steam.
Homework Equations
I get confused to exact what equation to use to figure out the energy required.
I *think* it would be this one:
Q_{t}=M_{water}C_{water}\DeltaT_{water} +...
Homework Statement
A child pushes a block of wood with a mass of 0.72kg across a smooth table. The block starts from a position of rest and after 2.0s it has a velocity of 1.6m/s [forward]. The coefficient of kinetic friction is 0.64.
I have already determined that
F_{net}= 7.06N
F_{f}=...
f=ma
14.55N = (0.165g)(a)
14.55/0.165g = (0.165g)(a)/0.165g)|
88.19m/s² = a
a=2(d)/t²
88.19m/s² = 2(d)/3²
88.19m/s²x9s = 2(d)/9 x 9
793.71m = 18d
793.71m/18 = d
44.095m
44.1m
Is that the correct way to do it??
Did I ever mention how much I dislike the hundreds of different...
Would the formula for finding the acceleration once the puck leaves the stick be:
a=v/t
a=45m/s / 3.0s)
a=15m/s²
so then would it be:
a=2(d)/t²
15=2(d)/9
15x9 = 2(d)x9
135 = 18d
135/18 = 18d/18
7.5m?
So in this case the puck would travel 7.5m.
Homework Statement
A resting 164 g hockey puck is hit with a force of 15.3N. The frictional force slowing the puck down is 0.75N.
a) Find the net forces acting on the puck. Remember to consider both horizontal and vertical forces
b) If the puck leaves the stick with a velocity of 45m/s...
Homework Statement
Two ropes are attached to a log that is floating in the water. A force of 80.0 N is applied to one rope and a force of 60.0 N is applied to the other rope, which is lying at an angle of 40° from the first rope. What is the net force on the log?
We know that the angle...