Specific Heat problems

[mustang]

Problem 7.
Given; specific heat of water=4186 J/kg* degrees Celcius
Brass is an alloy made from copper and zinc. A 0.66 kg brasss sample at 98.6 degrees Celciusus is dropped into 2.33 kg of water at 4.6 degrees Celcius. If the equilibrium temperature is 7.0 degrees Celciusus, what is the specific heat capacity of brass? In J/kg*Celciusis.

Problem 9.
A student drops metallic objects into a
224 g steel container holding 325 g of water
at 28±C. One object is a 472 g cube of copper
that is initially at 87 degrees C, and the other is a
chunk of aluminum that is initially at 9 degreesC.
To the students's surprise, the water reaches
a final temperature of 28 degreesC, precisely where
it started.
What is the mass of the aluminum chunk?
Assume the specifc heat of copper and alu-
minum are 387 J/kg degrees C
± C and 899 J/kg degrees C

Problem 12.
Given: specific heat of water = 4186 J/kg degrees Ceicuis
and water's latent heat of fusion = 3.33 *10^5 J/kg.

A 0.012 kg cube of ice at 0.0degrees Ceicuis is added to
0.459 kg of soup at 80.4degrees Ceicuis.
Assuming that the soup has the same specific
heat capacity as water, find the final tem-
perature of the soup after the ice has melted.
Answer in units of degrees Ceicuis.

[Cummings]

Rightyo. A handy thing to do for specific heat problems is to imagine "what is loosing heat" and "what is gaining heat"

So for Problem 7, the water gains heat and the brass looses heat

And, given that no energy is lost externaly (a simplicication we use for such problems) all the energy lost by the brass must equal the energy gained by the water.

the energy lost or gained is given by mc(Tf-Ti)
where
m is the mass of a substance,
c is the specific heat capacity of a substance,
Tf is the final temperature of the substance
Ti is the initial temperature of the substance

we use this for both the brass and water

Now, first law of Thermodynamics states that two objects will gain or loose heat to be in thermal equilibrium with each other. So Tf for both the water and the brass will be the same.
Brass = Water
mc(Tf-Ti) = mc(Tf-Ti)
.66 * c * (7-98.6) = 2.33 * 4186 * (7-4.6)
Now the only thing you need to know is c (specifiv heat of brass). So solve for c.

For Problem 9, the water and aluminum gain heat and the copper looses heat. The steel container is mentioned but no specific heat is given for it so i assume you are not to use it. If you are you have to take note that it also gains heat (and is at the same temp as the water due to thermal equilibrium)
so use water + aluminium = copper

For Problem 12, the soup looses heat and the ice gains heat going from 0 to equilibrium and also from changing state from solid to liquid.( this takes up l*m jouls where m is the mass of ice and l is the latent heat of fusion.)

[mustang]

Could you go more in dept with problem 9.

Water+ aluminium = copper

Water : m=325g
28 degrees Celciusus

aluminium: m=?
specifc heat: 899 J/kg degrees C

copper: m=472g; temp=87 degrees C
specifc heat: 387 J/kg degrees C