The question cannot be answered with the information given. You also
need to know the specific heat of Pb. The specific heat of a
substance is the ratio of the heat capacity of the substance to that
of water. The heat capacity of a substance is the amount energy
needed to raise the temperature of a unit amount (usually expressed in
molar or mass units) of the substance by 1 degree C or K.
Looking up the specific heat of Pb on Google yields a number of hits,
such as: http://environmentalchemistry.com/yogi/periodic/Pb.html. The
specific heat of Pb is given as 0.13. (Note that this website gives
the wrong units for specific heat. Specific heat should be unitless
because it is defined as the ratio of two quantities that have the
same units, so all the units cancel.)
Assuming 1) that the heat capacities of water and Pb are constant over
the temperature range of 20 - 92.5 C; 2) that no heat is lost from the
system (i.e., that the container is a perfect insulator); and 3) that
no mass is lost from the contents of the container (for instance by
evaporation of the water), we can use the law of conservation of
energy to say that the energy lost by cooling the piece of lead must
go into heating up the water. By definition, at equilibrium, the
temperatures of the lead and water must be equal.
In general, then, if T1 is the initial temperature of the water, T2 is
the initial temperature of the lead (or any other substance), T3 is
the equilibrium temperature, M1 is the mass of water, M2 is the mass
of the lead (or other substance), C1 is the heat capacity of lead (or
other substance), and C2 is the heat capacity of water:
C1*M1*(T1-T3) + C2*M2*(T2-T3) = 0
The first term on the left hand side of the equation is the energy
change due to the change in temperature of the lead, and the second
term is the change in energy due to the temperature change of the
water. By conservation of energy the sum of these changes must equal
zero.
Dividing through by the heat capacity of water, and noting that the
specific heat of lead is defined (see above) as S=C1/C2, we have:
S*M1*(T1-T3) + M2*(T2-T3) = 0
Solving for T3:
T3= (S*M1*T1 + M2*T2)/(S*M1 + M2)
Plugging in the values for the temperatures, masses, and specific
heat:
T3 = (0.13 * 14.9g * 92.5C + 20C * 165g)/(0.13*14.9g + 165g)
T3 = 20.84 C
One final esoteric caveat is that the lead in the sample has an atomic
mass of ~207.2, which is the nominal atomic mass for "common" lead.
In fact, the atomic mass of lead samples in nature can vary
consideably because the three most abundant isotopes of lead are the
final, stable decay products of the U and Th radioactive decay series.
(206Pb is the daughter of 238U, 207Pb is the daughter of 235U, and
208Pb is the daughter of 232Th. The only other stable isotope of lead,
204Pb, has no radioactive parents.) The isotopic composition (i.e.,
the relative abundances of the different isotopes) of lead varies
considerably in nature, depending on the time-integrated U/Pb and
Th/Pb ratios of the rocks that the lead originated in. This matters
because 14.9 grams of pure 208Pb would contain ~1% fewer atoms than
14.9 grams of pure 206Pb ((208-206)/208 ~ 0.01 = 1%). This will
affect the value of the heat capacity if the heat capacity is
expressed in mass units. |
