[ch101em]

A 1000 g shot of lead, at 300 degrees Celsius, is dropped into 100 g of water at 5.6 degrees Celsius.  The specific heats of lead and water are 0.129 and 4.184 J/(g*C).  What is the final temperature of the water?

100
65
40
75
50

Also

A 100.0 g sample of water at 80°C, is mixed with 10 g of ice cubes at 0°C. The heat of fusion of ice is 335 J/g, and the specific heat of liquid water is 4.18 J/g°C. Calculate the final temperature in degrees Celsius of the mixture.

0.0
10.6
30.7
65
53.2

Thanks

[enahs]

These all are based on the assumption that no heat is lost by the system.

That is,
Q₁ + Q₂ = 0
or
Q₁ = -Q₂

Where Q = m · C · ΔT

The second one is pretty much the same as the first, but you must account for the energy to change state (solid-liquid).

We are not here to do your work for you, but help you. Show your attempts and try with what I told you.

[ch101em]

OK, I get the fact that you have to set up equations, pretty cool.  Sorry for the troubles.

Let the final temperature be x.

Then (300-x)*1000*.129+(x-5.6)*100*4.184=0.  That is cool.

I just didn't know the fact that the sums of the heat capacities would equal 0.  Is there a specific reason why?

[enahs]

I just didn't know the fact that the sums of the heat capacities would equal 0.  Is there a specific reason why?

Yes. We makes the assumption that no energy in the form of heat is lost or gained by the system, that it is perfectly insulated. This means the total energy in the form of heat can not change magnitude.

This means for the temperature of one thing to increase, energy must flow into it, the energy has to come from somewhere, the system or the surroundings.

This system is better represented mathematically as:
Q_system = -Q_Surroundings
But that only works for two things, so move them to the same side and you are left with 0; so you can use the same assumption for a total system (system + surroundings) with more then 2 components.

[Ryder]

Let the final temperature be x.

Then (300-x)*1000*.129+(x-5.6)*100*4.184=0

In this eq why have we used (300-x) and not (x-300)? Is it because the temp is being lowered.

Also I had a similar question but when I used the above method then my final temp(119) came way above my initial temp and when I used (x-..) x came out to be 4.33.

Q. Suppose you mix 20.5g of water at 66.2C with 45.4g of water at 35.7C in an insulated cup. What is the max. temp. of the sol. after mixing?

I used the eq. (20.5)(4.184)(x-66.2)= -(45.4)(4.184)(x-35.7) to get 4.33
and                    "       "    (66.2-x)=     "       "           "       to get 119.59

Pls reply soon since I need to submit this today.

[ARGOS++]

Dear Ryder;

Maybe you have already realised that the final temperature must lie between 35.7°C and 66.2°C.

Hint: The sign in your Equ. 1 are not correct, but on both sides, so you must have done a mistake during rearrangement the Equ. 1 for isolating x on one side.
(For the correct signs re-read the first posting from enahs.  ΔT = T_start – T_end.)

Add. Hint: The spec. heat capacity for water you can delete as first, because it’s identically as factor on both side of the Equ.

Good Luck!
                    ARGOS⁺⁺