[UG]

I know this is a basic question and I feel really stupid asking it 

Where do I begin to work this out?

Calculate the final temperature resulting from placing 10 g of aluminium at 320^oC into 300g of water at 20^oC.
c for liquid water = 4.18 J^oC^-1g^-1
c for aluminium = 0.900 J^oC^-1g^-1

[cliverlong]

I know this is a basic question and I feel really stupid asking it 

Where do I begin to work this out?

Calculate the final temperature resulting from placing 10 g of aluminium at 320^oC into 300g of water at 20^oC.
c for liquid water = 4.18 J^oC^-1g^-1
c for aluminium = 0.900 J^oC^-1g^-1

1. You are placing a hot object (the aluminium) into (or in contact with) a cooler object - the water. What then happens?
http://en.wikipedia.org/wiki/Laws_of_thermodynamics#Zeroth_law

2. Look at the units for the specific heat capacity: J^oC^-1g^-1
which of those variables do you know for the aluminium and the water? What can you calculate from the data you have? How can you combine this value with the zeroth law to work out what change will ocur between the two objects? When will the change appear to have completed?

Clive

[UG]

1. You are placing a hot object (the aluminium) into (or in contact with) a cooler object - the water. What then happens?

The aluminium should lose energy (heat) and the water should gain the energy (heat) which is lost.

2. Look at the units for the specific heat capacity: J^oC^-1g^-1
which of those variables do you know for the aluminium and the water? What can you calculate from the data you have? How can you combine this value with the zeroth law to work out what change will ocur between the two objects? When will the change appear to have completed?

Okay, so I've got the mass in grams and the temperature so we need to find the energy that each of the objects have right?? Is there a specific equation? Is it ΔH=mcΔT?  But what is ΔT? Am I on the right track here?

[cliverlong]

1. You are placing a hot object (the aluminium) into (or in contact with) a cooler object - the water. What then happens?

The aluminium should lose energy (heat) and the water should gain the energy (heat) which is lost.

2. Look at the units for the specific heat capacity: J^oC^-1g^-1
which of those variables do you know for the aluminium and the water? What can you calculate from the data you have? How can you combine this value with the zeroth law to work out what change will ocur between the two objects? When will the change appear to have completed?

Okay, so I've got the mass in grams and the temperature so we need to find the energy that each of the objects have right?? Is there a specific equation? Is it ΔH=mcΔT?  But what is ΔT? Am I on the right track here?

Ok, almost there, I think.

a. So the hotter body loses heat to the cooler body (Zeroth law)
b. This nett exchange of energy will continue until the bodies are at the same temperature (Zeroth law)
c. The amount of heat transferred from the hotter to the cooler body is equal (ΔH) (conservation of energy)

So ... if the new equlibrium temperature attained by both bodies becomes X , where 20^oC < X < 320^oC

How do you think ΔT relates to the three temperatures: 20^oC, X, 320^oC ?

Regards,

Clive

[UG]

ΔT = (20-X) + (X-320) ?? 

[cliverlong]

ΔT = (20-X) + (X-320) ?? 

ΔH transferred = m_Alc_Al(320 - X) =  m_waterc_water(X - 20)

Substitute values, rearrange, isolate X.

Clive

[typhoon2028]

I love these problems.  I suppose you are to assume 320 oC aluminium will not flash any of the water?

[UG]

ΔT = (20-X) + (X-320) ?? 

ΔH transferred = m_Alc_Al(320 - X) =  m_waterc_water(X - 20)

Substitute values, rearrange, isolate X.

Clive

ΔH transferred = m_Alc_Al(320 - X) =  m_waterc_water(X - 20)

10 x 0.9 (320-X) = 300 x 4.18 (X-20)
          9(320-X) = 1254(X-20)
           2880-9X = 1254X-25085
              27960 = 1263X
                     X = 22.138 ? 
So ΔH transferred is 22.138 Joules ? 

[UG]

Or is that meant to be ΔT = 22.138 ^oC

[cliverlong]

Or is that meant to be ΔT = 22.138 ^oC

I haven't checked your algebra, but seems fine to me.

Is this sensible?

Well we have a small amount of aluminium relative to the amount of water and the heat capacity of water  is large relative to the aluminium so even though the Al is hot it causes a relatively small rise in the temperature of the water as the Al cools.

I agree with the post above that plunging 320^oC Al into water would probably cause the water that first came into contact with the metal to boil - so the calculation is probably unrealistic.

Clive

[UG]

Okay, thank you very much 

[amir]

I know this is a basic question and I feel really stupid asking it 

Where do I begin to work this out?

Calculate the final temperature resulting from placing 10 g of aluminium at 320^oC into 300g of water at 20^oC.
c for liquid water = 4.18 J^oC^-1g^-1
c for aluminium = 0.900 J^oC^-1g^-1

Hi, see there is a simple equation that u should use it:
Q=m*c *deltaT          IN THIS EQUATION: Q=transfered heat or energy (J) , m= mass of substance (Kg),
                                                        delta T= T1 -T, T1: initial temperature for any        substance and T is the final temperature for any substance in thermal equilibrium


now, u can solve the problem, see: T1(Al, 10 gr)=320 c, T1(H2O, 30 gr)=20 c,
                                                 c (Al) =0.9           , c (H2O)=4.18,      we should calculate T=?

Al is warmer than H2O so the heat transfers from Al to H2O, this is Q, so the amount of decreasing energy in Al=amount of increasing energy in H2O, We can write fundamental equation in this form      m(Al) c(Al) ( T1(Al)-T)=m(H2O) c(H2O) ( T1(H2O)-T)
 notice: equilibrium temperature for both substance that are in thermal equilibrium are the same = T
 now, just replace the quantities of any variables(m,c,T1) and calculate T.for further information u can see physical chemistry books. hope you got it.

ALL THE BEST,
Amir