CU PHYS221 Practice Test 3 Solution - vC

# Find t f know t ibullet 3270 ºc m bullet 00095 kg c

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Find: T f Know: T i(bullet) = 327.0 ºC m bullet = 0.0095 kg c Pb = 128 J/kg-ºC T i(water) = 20.0 ºC m water = 0.10 kg c water = 4,186 J/kg-ºC Sketch: Equation(s): Q = mc (T f −T i ) Q lost + Q gained = 0 Solve: This is a thermal equilibrium problem 9 Version C

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PHYS221 Practice Test 3 If there are no losses to the system, the heat lost by the 0.0095 kg lead bullet will equal the heat gained by the 0.10 kg of water or Q lost + Q gained = 0. Both the lead bullet and water will have the same final temperature “T f ” at thermal equilibrium. Mathematically, the heat lost by the 0.0095 kg lead bullet: Q lost = m bullet c Pb (T f − T i(bullet) ) The heat gained by raising the temperature of the 0.10 kg of water is: Q gained = m water c water (T f − T i(water) ) At thermal equilibrium: Q lost + Q gained = 0 Substitute in the two expressions for heat: m bullet c Pb (T f − T i(bullet) ) + m water c water (T f − T i(water) ) = 0 Solve for “T f m bullet c Pb T f − m bullet c Pb T i(bullet) + m water c water T f − m water c water T i(water) = 0 Collect common “Tf” terms m bullet c Pb T f + m water c water T f = m bullet c Pb T i(bullet) + m water c water T i( water) Factor out common “T f ” terms T f {m bullet c Pb + m water c water } = m bullet c Pb T i(bullet) + m water c water T i(water) Solve for “T f T f = m bullet c Pb T i(bullet) + m water c water T i(water) {m bullet c Pb + m water c water } Substitute in known values “m bullet = 0.0095 kg”, “c Pb = 128 J/kg-ºC”, “m water = 0.10 kg”, “c water = 4,186 J/kg-ºC” “T i(bullet) = 327.0 ºC”, “T i(water) = 20.0 ºC”, and solve for “T f T f = {(0.0095 kg)(128 J/kg-ºC)(327.0 ºC) + (0.10 kg)(4,186 J/kg-ºC)(20.0 ºC)} {(0.0095 kg)(128 J/kg-ºC) + (0.10 kg)(4,186 J/kg-ºC)} T f = 20.9 ºC 10 Version C
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