# Unknown specific heat of a mixture of several substances of known specific heat

1. Mar 5, 2011

### torquemada

1. The problem statement, all variables and given/known data

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/°C · g) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J/°C · g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 96.5 g of water at 23.5°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

2. Relevant equations

(the specific heat of a substance *Joules)/(degrees C * # of grams) = 1

3. The attempt at a solution

I first found the Energy in Joules in all 3 substances. For Al I got 561.797J, Fe 2222.22J, H2O 542.5239J

Then for the entire mixture the total J is 3326.5409 and the Mass total is 111.5g, and I need to find Temperature. I'm stuck at trying to find the Specific heat of this mixture. In my first attempt I got 3.6979 by adding up the specific heats of each substance by % of Total Mass. Then with that I calculated Temperature and got around 110 degrees C which was wrong according to WebAssign. My second attempt I just used the Specific heat of water 4.18 and I got around 125 C which was also wrong. I am leaning toward the first method to get Significant heat but maybe my calcs are off or I rounded wrong somewhere because WebAssign doesnt really have flexibility when it comes to rounding/sig figs. Or maybe a fundamental aspect of my approach to this problem is wrong? Just a shot in the dark but I tried adding the specific heats without regard for the mass % and I got 5.52 which gave me a Temperature of 164.7 C - i have a feeling this is an incorrect approach but I'm just throwing it out there so you know I thought of it. Thx in advance for any help

Last edited: Mar 5, 2011
2. Mar 5, 2011

### Dadface

Hi torquemada.Let the final temperarure=T.How much heat will be lost by the Al when the temperature drops from 100 to T?how much heat will be lost by the iron?What gains these heat losses?

3. Mar 5, 2011

### torquemada

the heat lost for aluminum will be (100-T)g/.89. the heat will go to water because no heat is lost to the surroundings. Same for Iron except using .45. But T is unknown so what does that do for us?

Last edited: Mar 5, 2011
4. Mar 5, 2011

### Dadface

Heat = mass*shc*temperature change so for aluminium the heat loss is given by:
H=5*0.89(100-T)
Similar for iron but the shc and mass are different.
The water rises in temperature from 23.5 to T so write an expression for heat gained.
(I think the shc of water is 4.2)

5. Mar 5, 2011

### torquemada

I got 96.5g*4.18(23.5 + T) = 5g*.89(100-T)+10g*.45(100-T)

and I got -20.819 for Delta T. What is the original temperature though? I need to subtract Delta T from the original temperature which can't simply be 100 Celsius because the 2 substances of Iron and Aluminum were both at 100 Celsius, then they were first mixed together before being thrown in the water so wouldnt that yield a higher temperature? Or does one simply go with 100 celsius and if so what is the reasoning behind that? When I calculated the temperature of the iron/aluminum mixture prior to being thrown in the water I got 110.742 degrees Celsius. Basically the conceptual question I'm asking is if two different substances with different heat capacities are both at the same temperature and they are mixed, does the temperature change? And if it does change it probably is dependant on the masses of each substance no?

Oops or maybe I'm reading the question wrong - maybe it reads that after they are mixed the mixture is 100 Celsius. I was reading it as they are each individually heated to 100 each and then mixed together. Not sure if that would make a difference or not though, as per my question above.

Last edited: Mar 5, 2011
6. Mar 5, 2011

### Dadface

The water rises in temperature so the heat gained is given by:
96.5*4.18(T-23.5).
By equating heat gained to heat lost you should be able to finish.

7. Mar 5, 2011

### torquemada

Ok thanks. I guess my next question is how is this problem different from A 32.0-g sample of water at 290. K is mixed with 51.0 g water at 300. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

For this problem I was able to find the final temperature just by using

(the specific heat of a substance *Joules)/(degrees K * # of grams) = 1

I found each internal energy separately, then added up the energies and masses and got the new temperature using the above equation. Is that only working because its the same substance with the same heat capacity?

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