Lead pellets, each of mass 0.50 g, are heated to 200°C. How many pellets must...?
Lead pellets, each of mass 0.50 g, are heated to 200°C. How many pellets must be added to 520 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C? Neglect any energy transfer to or from the container.
- DHLv 710 years agoFavorite Answer
Q out of the lead = Q into the water
Q in = m*c*ΔT = 520g*4.186J/g-oC*5oC = 10884J
So for the lead m = Q/(c*ΔT) = 10884J/(0.130J/g-oC*175) = 478.4g
So you must add 478.4/0.50 = 956.8 (rounded up to 957 pellets)
- 4 years ago
q = cmdT qgain = qlost c1mdT (water) = c2mdT (lead) dT (water) = 25 - 20 dT (lead) = 200 - 25 m (lead) = 490 g m (water) = unknown c1 = water c2 = lead c for water and lead are constants, but I don't remember them, so you look them up. m(water) = (c1/c2) * (dT(lead)/dT(water)) * m(lead) m(water) / 0.80 g = # pellets