Coffee cup calorimetry.

Styrofoam cup calorimetry?

• A sample of Pb (specific heat = 0.128-J/g°C) weighing 110.7-g is boiled in water at sea level until it reaches 100.0°C. The sample is dumped into a coffee cup containing an unknown mass of water that had reached equilibrium temperature 23.5°C. The temperature of the water and Pb rose to 25.3°C. Assuming that no heat is absorded by the coffee cup, what was the mass of the water in the coffee cup. A sample of Hg (specific heat = 0.139-J/g°C) weighing 230.9-g is boiled in water at sea level until it reaches 100.0°C. The sample is dumped into a coffee cup containing 119.7-g of water that had reached equilibrium temperature 23.5°C. The temperature of the water rose. Assuming that no heat is absorded by the coffee cup, what was the final temperature in°C?

• Answer:

  i just learned about this stuff- lucky you huh? :P q of system (system= whats being added into water which happens to be Pb) = q of surroundings (which in the case is the water- but it could be anything outside of the cup too). therefore: q (mcdeltaT) of Pb = q (mcdeltaT) or H20 Stuff thats given to you in question you should write out first- that way you can just add all the stuff into the equation: CsPb= 0.128J/g°C mass Pb= 110.7g delta T of Pb= 100°C DeltaT solution= 25.3°C Cs H20= 4.18J/g°C Delta T= 23.5°C mH20= ?? ok now add this all into it. mcdeltaT of Pb = mcdeltaT of H20 110.7gx0.128Jx(100-25.3)°C g°C = 1058.47 J of Pb now do the same for H20 m x 4.18J x (25.3-23.5)°C g°C = 7.524 then mass H20= 1058.47J divided by 7.524 140.38 g therefore, assuming no heat was absorbed, 140.38 g of water was in the coffee cup. i have to go so, im hoping from this example that you understand and can do the next one. :)