Quantity of Heat/Specific Heat Water+Iron

  • Thread starter Thread starter Kaleb
  • Start date Start date
  • Tags Tags
    Heat
AI Thread Summary
The discussion centers on calculating the final temperature when 100g of water at 20°C is mixed with 100g of iron nails at 40°C. The key equations involve equating the heat lost by the iron to the heat gained by the water, using their specific heat capacities. The initial confusion arises from misidentifying initial and final temperatures in the calculations. After clarification, the final temperature is determined to be 28°C, confirmed by solving the equations correctly. The thread emphasizes the importance of understanding the heat transfer principles involved in the problem.
Kaleb
Messages
47
Reaction score
0
[SOLVED] Quantity of Heat/Specific Heat Water+Iron

Homework Statement


Exact Problem:

What will be the final temperature of 100g of 20 degrees C water when 100g of 40 degrees C iron nails are submerged in it. (The specific heat of iron is .12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails.)


Homework Equations



Q = cm(change in)t

Q = quantity of heat

c = specific heat
(water = 4.184J/g or 1 cal/g)
(iron = .502J/g or .12 cal/g)

m = mass
100g + 100g = 200g total substance(water+iron)

t = temperature
(40(final)-20(initial))

The Attempt at a Solution



I feel I slightly understand what I am supposed to do but the part where it says I need to find the heat lost by iron and the heat gained by water I am totally confused about. Heres my attempt: My assumption is that I need to do two different calculations and difference them out. Also if someone could point me in the right direction to be able to post real equations instead of what I have shown that would be so awesome.

Iron First:
Q = cm(Tf-Ti)

Q = .502J*100g(40-20)

Q = 1004J of Energy Lost

Water Second:

q = 4.184*100g(40-20)

Q = 8368J of Energy Gained

Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g

I have tried numerous ways to find the temperature difference, lost and gained but I am clueless. Also I can't figure out where to transfer back into Celcius. By my calc's it would have gained like 1700 degrees haha. >.> I feel dumb... Anyways TIPS are always appreciated in advance. Constructive critism is appreciated as well, and well anything else that can make me a better poster is welcome too. ^^ thanks!
 
Physics news on Phys.org
Your only problem appears to be that you have not read the problem correctly and don't seem to know what is being asked! Your start is correct but why are you using 40 degrees and 20 degrees as "initial" and "final" temperatures for both water and iron?

You also say
"Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g "
which you should realize makes no sense. The question was "What will be the final temperature" and that's not even a temperature.

20 degrees is the initial temperature for the water and 40 degrees is the initial temperature for the iron. The final temperature of both will be the same, of course, so just call that "T".

Your equations should be
Iron: Energy lost Q = .502J/g*100g(T- 40) and
Water: Energy gained q = 4.184J/g*100g(T-20) and they must be equal. Solve for T.
 
I see the fault, Thanks a bunch Ivy. You have saved my day and made me feel all the more dumb O.o. That was a joke btw.

Edit: I came out with 32 degrees C as the final for both. Any chance I can get confirmation on this? Ill wait awhile and if nobody posts Ill mark as solved. Thanks again Ivy, your my hero!
 
Last edited:


Oh my GOD

Temperature will be 28 C in equilibrium with water and nail.
If you have eqations ready is simple.
Put them L one = P one and find answer for T.
Nails will lose from 40C - 12C= 28C
Water will gain heat from 20C + 8C = 28C also
very simple and time consuming but some people are fast so it is their own negative to be not able to stop and think for a while to grab the answare in the balls
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Egg Drop Challenge! Need ideas for winning designs'

I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
View full post »

Similar threads

I try to solve a thermodynamics problem on heat transfer
Replies
3
Views
2K
Problem involving specific heat and graph
Replies
3
Views
1K
The Mystery of Heat Loss: Examining Ice Melting Without Heat Loss
Replies
11
Views
2K
Specific Heat Question: Calculating Temperature Change with Copper and Water
Replies
2
Views
1K
Do you need to account for the water heating up to the boiling point?
Replies
23
Views
2K
Specific Heat Problem, not getting the correct answer
Replies
5
Views
2K
Calculate Resultant Temperature: 100g Steam at 100°C + 500g Ice at -20°C
Replies
1
Views
1K
Thermal Equilibrium of Hot Iron & Water
Replies
5
Views
2K
How much ice to cool down this water?
Replies
17
Views
5K
Problem with heat capacity of thermal system
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top