In the last Basic Physics instalment we left off with a series of problems:

1) Work is done on water by a rotating paddle wheel driven by two weights of 1 kg mass each, falling at constant speed. Water is then stirred by the paddles increasing its temperature. Assuming that the loss in potential (mechanical) energy = work done by the paddle wheel on water, where the latter is defined according to:

m(H2O) c(H2O) (T2 - T1), and T2 = 15.5 C, T1 = 14.5 C,

find the height needed for the weights to drop to obtain 4.18J. (take the specific heat of water, c(H2O) = 4200 J/Kg K, and g = 9.8 m/sec/sec and the mass of the water = 10 kg.

2) A 200 pound weight falls 40 feet. How much heat was developed? (Take g = 32 ft/sec/sec and 1 Btu = 778 ft-lbs.)

3) A lead bullet (mass m) traels at velocity v = 300 m/s, and is fired into a wood target. Assuming that half the heat developed goes into the target and the rest into the bullet, how much will the temperature of the bullet be raised?

4)A 2 kg object has a speed of 2.5 m/s. Find: (a) its kinetic energy in joules, and b) the equivalent energy in calories.

time

The solutions are as follows:

(1) Find work done by paddle wheel on water:

m(H2O) c(H2O) (T2 - T1) = (100 kg)(4,200 J/kg K) (1 K) = 4.2 x 10^5 J

But we need to find the height dropped (h) to obtain nearly 100,000 time less (e.g. 4.18J). So: Loss in potential energy of falling weights = energy gained = 4.18J, or 2 mgh = 4.18 J = 2 (1 kg) (9.8 m/sec/sec) h, so that:

h = (4.18J)/ (19.6 kg m/sec^2)= 0.21 m.

(2) We let Q (amt. of heat) = mgh = (200 lb.) (40 ft.) = 8,000 ft-lb. (N.B. The value of g isn't needed, since lbs is already = mg (weight) so is already incorporated).

The amount of this Q in Btu = (8,000 ft-lbs)/ 778 ft-lb/Btu) = 10.3 Btu

(3)The kinetic energy of the bullet = E_k = ½ mv^2 = ½(m)(300m/s)^2

E_k = 45 000m (J) and this is equal to:

4.5m x 10^4 J

Let t be amount the temperature is raised and the specific heat of lead = 128 J/kgK

Then: (128 J/kg K)(m) t = ½( 4.5m x 10^4 J) = 2.25 x 10^4 J

Canceling 'm' from both sides and equating:

128 J/kg K(t) = 2.25 x 10^4 J or t = (2.25 x 10^4 J)/ (128 J/kg/K) = 175 K or 175C (Since the Kelvin temperature difference is equivalent to the Centigrade one)

(4)(a) KE in Joules = ½ m v^2 = ½ (2 kg)(2.5 m/s)^2 = 6.25 J

(b) the equivalent energy in calories:

Since by Joule's law (J = W/H)so 1 cal = 4.18 J, then

6.25 J = (6.25J)/(4.18 J/cal) = 1.5 cal

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