You can find the specific heat of lead (C1) here:
http://www.sciencebyjones.com/specific_heat1.htm
The specific heat of water is
C2 = 1.000 Cal/g C
Set the heat gained by the water equal to the heat lost by the lead:
M1*C1*(92.5 - T)
= M2*C2*(T-20.0)
Now plug in the appropriate values for the masses M and the specific heats C, and solve for T.
M1 = 14.9 g
M2 = 165 g
Can someone check my answer for me...
1)Suppose a piece of lead with a mass of 14.9 g at a temperature of 92.5 degrees 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. What is the final temperature of the system?
Answer: 20.19 degrees
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