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A bullet made of solid gold has mass m = 10 grams. It is shot from a rifle at v = 500 meters per second. As it imbeds itself in a plastic target, one-half of its kinetic energy is turned into heat.

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Question 1:
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What is the kinetic energy of the bullet?
How much energy is turned into heat?

Answer: The kinetic energy of the bullet is KE = 0.5*m*v^2 = 0.5*(0.01 kg)*(500 m/s)^2 = 1250 J If one-half of this KE is turned into heat, then heat energy = 0.5*(1250 J) = 625 J

(If you don't know how to do problem 1, assume that the amount of heat energy is 1000 Joules).

Assume that all the heat energy acts to raise the temperature of the bullet, which increases from T = 20 Celsius to T = 437 Celsius.

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Question 2:
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What is the specific heat capacity of gold?

Answer: The specific heat capacity of gold relates the heat energy added to the bullet to its change in temperature: heat added = (specific heat)*(mass)*(change in temp) We can solve for the specific heat capacity of gold: heat added specific heat = ------------------------- (mass)*(change in temp) 625 J = ------------------------------- (0.01 kg)*(417 Celsius degrees) = 150 J/(kg*Celsius degree)

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Question 3:
**Convert the final temperature of the bullet from
Celsius to Fahrenheit, and to Kelvin.

Answer: Bullet is at T = 437 degrees Celsius. temp Kelvin = (temp Celsius) + 273 K = 437 + 273 K = 710 K temp Fahr = (temp Celsius)*(9/5) + 32 = (437)*(1.8) + 32 = 819 degrees F

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This page maintained by Michael Richmond.
Last modified Dec 19, 1997.
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Copyright © Michael Richmond. This work is licensed under a Creative Commons License.