calorimetry......please help?

Question

1. a 3.52 kg lead can inside a calorimeter contains 2.25 kg of water at 37.5 degree celsius.What mass of silver at 108.6 degree celsius added to the calorimeter would result in a final temperature of 45 degree celsius.
(C of water- 4186 J/kg-Cdegree, C of lead- 128 J/kg-Cdegree, C of siver 235 J/kg-Cdegree)

2.A combination of 0.250 kg of water at 20 degree celsius, a 0.400 kg of aluminum at 26 degree celsius and 0.100 kg of copper at 100 degree celsius is mixed in an insulated container and allowed to come to thermal euilibrium.Ignore any energy transfer to or from the container and determine the final temperature of the mixture.( C of water-4186 J/kg-Cdegree, Caluminum -910 J/kg-Cdegree, C of copper- 390 J/kg-Cdegree)

Answer 1

Q = mcDT or more to the point its the sum of the contributions:

Assume the lead is at the same temp as the water. Then:

(mPb*cPb+mH20*cH20)*(45 - 37)=mAg*cAg*(108.6-45)

The heat gained by the lead and water has to come from the silver.

mAg = (mPb*cPb+mH20*cH20)*8/(cAg*63.6)

= (3.52*128+ 2.25*4186)*8/(235*63.6) = 5.29 kg

For #2

Initially you have:

mH2O*cH2o*TH2O + mAl*cAl*Tal+ mu*cCu*TCu = (mH2O*cH2O + mAl*cAl+ mu*cCu)*T

So

T = (mH2O*cH2o*TH2O + mAl*cAl*Tal+ mu*cCu*TCu)/(mH2O*cH2O + mAl*cAl+ mu*cCu)

You can put the numbers in.

Answer 2

The only help you need is to figure out how to read your textbook and learn how to do these problems yourself. The concept is pretty straightforward. Heat is transferred from the hotter components to the cooler components until all components are at the same temperature.

In problem 1, the lead and water are the cooler components and the silver is the hotter component. Let the initial temperatures be TH for silver, TC for water and lead, and TM is the final temperature for all.
So: Heat lost by hotter component (silver)=
W(silver) * C(silver) * (TH-TM)
Heat gained by cooler components=
W(water) *C(water) * ( TM-TC) +
W (lead) * C(lead) * (TM-TC)
We know all the temperatures, we know all the C values, we only need to know the W(silver)

In problem 2, we have two cool components and 1 hot components, although the cool components start out with different temperatures (for some strange reason). You can set it up the same, allowing for the 2 different cool temperatures- you have all the masses and all the C values and all the starting temperatures.

SO DO IT!!