Suppose a piece of lead with a mass of 14.9 g at a temperature of 92.5
degrees C is dropped into an insulated container of water. The mass of
water is 165 g and its temperature before adding the lead is 20.0
degrees C. What is the final temperature of the system?

Hello teatea,

To solve this problem, we must use the principal of specific heat. 
This states that a change in heat is equal to mass*specific heat of
substance*change in temperature, or dQ = m*c*dT.  Since both parts of
the system will be changing the same amount in heat, we set the dQ of
each part of the system equal.

To use this formula, though, we must use kilograms as our unit of
mass. This means that the lead is .0149 kg and the water is .165 kg.

The specific heat (c) of lead is 125 J/kg*K and the specific heat of
water is 4200 J/kg*K (found here:
http://www.ex.ac.uk/cimt/dictunit/notes5.htm by searching for "specfic
heat lead").

The temperatures must also be converted to Kelvins as required by the
equation.  This means we must simply add 273 to each Celsius measure,
so the temperature of lead is initially 92.5 + 273 = 365.5 degrees K
and the initial temperature of water is 273 + 20 = 293 degrees K.

We now set up the equation as follows:

Q_lead = -Q_water

(.0149)(125)(T_f - 365.5) = (.165)(4200)(293 - T_f)

We solve this equation for T_f (final temperature), and find that it
is 293.19 K, or 20.19 degrees Celsius.

Please let me know if you require a clarification before rating the
solution.

Thanks,
chis-ga