also, i had a couple other questions i can't seem to really determine how to do

A cube of ice is taken from the freezer at -8.9掳C and placed in a 104 g aluminum calorimeter filled with 275 g of water at room temperature of 20掳C. The final situation is observed to be all water at 16掳C. What was the mass of the ice cube?

A nuclear power plant operates at 80% of its maximum theoretical (Carnot) efficiency between temperatures of 650掳C and 350掳C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?

if anyone has any suggestions i'd be extremely grateful!Physics help please!.. The 1.20 kg head of a hammer has a speed of 7.5 m/s just before it strikes a nail...?

For the first one, find the kinetic energy of the hammer right before it hits the nail using KE=1/2mv^2. (You should get 33.75 J)

Since the nail absorbs all the KE, we can assume it converts the KE into heat, so Q (heat energy) = 675 J. Now, since we know that Q=mc(change in T), (m- mass, c- specific heat capacity of iron, change in T- Temperature change) You can set up the equation,

675 J=(0.14 kg)(449 J/kg.K)(change in T).

Simplifying, (by dividing 675 by the product of .014 kg and 449 J/kg.K), we get that the nail's temperature goes up by 107.38 K, which is the same as 107.38 掳C.

FINALLY, WITH SIGNIFICANT DIGITS, WE GET THAT THE NAIL GOES UP BY 110 K.

For the second problem, you know the final temperatures of the water and the ice are the same (both 16 deg C). Because of this, we know that the heat exchange was also equal (that is, the water lost the same amount of energy the ice gained). So, we know that Q_1=Q_2.

This means:

mc(ch. T)_1=mc(ch. T)_2

(ch. T for ice is 16 - (-8.9) = 24.9 K

Setting up the equation,

m(2060 J/kg.K)(24.9 K)=(0.275 kg)(4180 J/kg.K)(4 K)

Simplifying, (multiplying everything on the right together and everything on the left together, then dividing the right by the left) we get that m=0.08964 kg, or 89.64 g.

USING SIGNIFICANT DIGITS, WE GET THAT THE MASS OF ICE WAS 90 g, or 0.090 kg.

As for the last problem, you must first find the efficiency of the power plant. using the equation

e=(hot-cold)/hot, you get that the ideal efficiency is about 46 %.

Since it is at 80 % ideal efficiency, multiply 46 by 0.8, leaving you with 37% efficiency. THIS WILL BE IMPORTANT LATER!!

Now, you know that the power is 1.3 GW, and power = work/second. This means that the energy PER SECOND is 1.3 GJ, or 1.3 billion J. To find how many that is per HOUR, multiply it by the number of seconds in an hour (3600). This leaves you with 4680 GJ per hour.

But you're not done yet. We've foun how much energy the plant IS using. We need how much it IS NOT using. To do this, we must find the total amount of energy produced per hour (we only found how much USEFUL energy it produced). Now, we use the efficiency again. we know that the amount of useful energy is equal to the efficiency (0.37) times the total amount of energy. So,

4680 GJ=0.37(TOTAL)

Simplifying, we get that the total is 12675 GJ. (NO, we're still not done)

To get your answer, we need to find how much of the total is not being used, so we subtract from the total the part that is being used.

12675 GJ - 4680 GJ = 7995 GJ

FINALLY, USING SIGNIFICANT DIGITS, WE CAN CONCLUDE THAT 8000 GJ OF HEAT IS DISCHARGED PER HOUR FROM THE POWER PLANT.

Hope you understood that! Good luck the rest of the year in physics!Physics help please!.. The 1.20 kg head of a hammer has a speed of 7.5 m/s just before it strikes a nail...?

You'll want to use Kelvin for the reactor plant efficiencies though.

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