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

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
HW 11 # 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
6 Q = mC (T-T)
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
# 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.)
Background image of page 6
*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.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 10/18/2010 for the course PHY 121 taught by Professor Stephens during the Fall '08 term at SUNY Stony Brook.

Page1 / 8

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

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online