Here are some helps on the frequently asked questions on this assignment:
Going back a few days in class, this question is asking you to find the Qf, or the energy of fusion for water as it melts. Unlike the problem we did in class this one is easier since your temp is a constant 0 degrees. This means that water can exist as ice at zero degrees or as liquid water…depends on the internal energy which is what you are to calculate in this problem.
Qf = m*Hf, where Qf is the energy of fusion and Hf is called the “heat” of fusion.
In this case you simply need to multiply the mass by the Hf (which you should google) to get the energy you are looking for.
This problem is VERY similar to the more difficult problem in #27. The only difference is that here you have to account for both the energy of fusion (Qf) and the energy from the change in temperature (mcdeltaT), so for 27:
Etotal = Qf + mcdeltaT
Etotal = mHf + mcdeltaT
Solve for Etotal
Number 28 is similar but involves that vapor state, so instead of Qf=mHf, you will need Qv=mHv and need to look up the heat of vaporization.
Using conservation of energy
KE = Thermal
.5Mv^2 = mcdeltaT, M = mass of car, m = mass of brakes Watch your units to ensure the right things cancel (kilograms in particular in this case).
Solve for deltaT.
Easy problem, similar in concept to #21, but difficult dimensional analysis to make it work out correctly:
KEbullet = mc(delta)T for the bullet
No pause and reflect with me for a moment on the silliness of the situation as described in the problem. What would really happen if a lead (soft) bullet strikes a steel (hard) plate? The bullet disintegrates into fragments and lots of KE and thermal energy are absorbed by the steel plate. However, in this problem the bullet apparently just stops at the point of impact and all the KE turns to heat. OK, so let’s go with that:
KE = mc(delta)T
.5mv^2 = mc(delta)T cancel m’s (interesting…)
.5v^2 = c*deltaT
Solve for deltaT. Easy right?
Careful! You have to match up your units! right now we have:
.5 = no units, v = m/s, c = Joules/(mass-degree) and T=degree
So, degrees cancel
(m/s)^ cancels with (m/s)^2 from the Joule
So in order to cancel the kg that is inside the Joules you must make sure your specific heat (c) has unit of…?
Correct, kg in the denominator.
Be careful, be wise, good luck!