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# hw 11 - The water is the hotter object Q_water =...

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HW 11 # 2

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6 Q = mC (T-T)

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# 7 #8 Given: density of water = 1000 kg/m^3 and specific heat of water = 4186 J/kg · ◦C. In a showdown on the streets of Laredo, the good guy drops a 15.4 g silver bullet, at a temperature of 17◦C, into a 194 cm^3 cup of water at 40◦C. Simultaneously, the bad guy drops a 15.4 g copper bullet, at the same initial temperature, into an identical cup of water. What is the coolest final temperature? (In other words, which one ends the showdown with the coolest cup of water in the West)? Neglect any energy transfer into or away from the container and assume the specific heat of silver and copper are 234 J/kg · ◦C and 387 J/kg · ◦C. (Answer in units of ◦C.)
*Round your answer please, thanks* Use the fundamental rule of Calorimetry (conservation of energy) Heat lost by hot object = heat gained by the cold object. This assumes that the experiment vessel is well insulated and no mass escapes.

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Unformatted text preview: The water is the hotter object: Q_water = m_w*c_w*(Tw1 - T2) The metal bullet is the colder object: Q_metal = m_m*c_m*(T2 - Tm1) Equate as per fundamental rule: m_w*c_w*(Tw1 - T2) = m_m*c_m*(T2 - Tm1) Solve for T2: T2 = (m_w*c_w*Tw1+m_m*c_m*Tm1) / (m_w*c_w+m_m*c_m) Express m_w in terms of its volume: m_w = rho_w*V_w T2 = (rho_w*V_w*c_w*Tw1 + m_m*c_m*Tm1) / (rho_w*V_w*c_w + m_m*c_m) Common data: Tw1:=40 C; Tm1:=17 C; rho_w:=1000 kg/m^3; c_w:=4186 J/kg-C; V_w:=194e-6 m^3; m_m:=0.0154 kg; Notice the unit conversion on V_w from cm^3 to m^3? Another unit conversion is used on the bullet masses from grams to kg. Data for the good guys: c_m:=234 J/kg-C; Good guy result: T2 = 39.898 C Data for the bad guys: c_m:=387 J/kg; Bad guy Result: T2 = 39.832 J/kg Conclusion: Bad guys win for using a bullet with a higher heat capacity, thus "weighting" the temperature more in favor of the bullet....
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hw 11 - The water is the hotter object Q_water =...

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