A piece of solid lead weighing
43.2 g at a temperature of 314 °C
is placed in 432 g of liquid lead
at a temperature of 367 °C. After a while, the
solid melts and a completely liquid sample remains. Calculate the
temperature after thermal equilibrium is reached, assuming no heat
loss to the surroundings.
The enthalpy of fusion of solid lead is
ΔHfus = 4.77 kJ/mol at its melting
point of 328 °C, and the molar heat capacities for
solid and liquid lead are Csolid =
26.9 J/mol K and Cliquid =
28.6 J/mol K.
Atomic mass of lead is 207.2 g/mol.
Mass of solid is 43.2 g
Number of moles of solid is
Mass of liquid is 432 g
Number of moles of liquid is
Let T deg C be the temperature after thermal equilibrium is reached.
Temperature change of solid is
Temperature change of molten solid is
Temperature change of liquid is
Heat lost by liquid
Heat gained when solid lead is heated from 314 deg C to 328 deg C (melting point)
Heat gained when molten lead is heated from 328 deg C(melting point) to T deg C
Heat gained when solid lead melts at 328 deg C (melting) point
heat lost by liquid heat gained by solid
Get Answers For Free
Most questions answered within 1 hours.