lilai3
  • lilai3
WILL REWARD MEDAL/FAN!!!((: A blacksmith heat an iron bar to 1445 Celsius. The blacksmith then tempered the metal by dropping into 42,800 cm^3 of water that had a temperature of 22Celsius. The final temperature of the system was 45 Celsius. What was the mass of the bar?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
so the temperature of the water has risen by \(45-22 = 23^{\circ}C\)
lilai3
  • lilai3
that's the delta t right?
ganeshie8
  • ganeshie8
Yes, then the heat added to the water is given by, \(Q=c*m*\Delta t=4.18*42800*23 = 4114792 J\)

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More answers

ganeshie8
  • ganeshie8
That means the bar has lost the same amount of heat. You can use that to find the mass of bar
lilai3
  • lilai3
Oh I see.
lilai3
  • lilai3
so basically it will be 4114792 = mass x 0.446 x (1445-45) Celsius which gives you the mass is 65890 grams or 6.6 kilograms?
ganeshie8
  • ganeshie8
0.446 is specific heat of iron is it ?
lilai3
  • lilai3
yeah
ganeshie8
  • ganeshie8
Looks good !
lilai3
  • lilai3
thanks can you also help me with this one?
lilai3
  • lilai3
a 752 cm^3 sample of water was placed in a 1 kg aluminum pan. The initial temperature of the pan was 26 Celsius, and the final temperature of the system was 39 Celsius. What was the initial temperature of the water?
ganeshie8
  • ganeshie8
looks like a similar problem.. give it a try
lilai3
  • lilai3
well i got stuck because i wasn't sure 0.902 j/ 1gC x (39- delta t) x 1000g = 752 cm^3 totally got that wrong i'm guessing
ganeshie8
  • ganeshie8
First, write down the given info Water : \(m = 752 cm^3 = 752 g\) \(c = 4.18 J/G^{\circ}C\) \(\Delta t = ?\) Aluminium pan : \(m = 1kg= 1000 g\) \(c = 0.9 J/G^{\circ}C\) \(\Delta t = 13\)
lilai3
  • lilai3
how do we know what the delta t is if we don't have the q of the equation ( i think that is the energy released)?
ganeshie8
  • ganeshie8
Energy released by water = Energy absorbed by aluminium pan
ganeshie8
  • ganeshie8
\[4.18*752*\Delta t = 0.9*1000*13 \] you can solve \(\Delta t\)
lilai3
  • lilai3
oh okay and would that be 39-26?
lilai3
  • lilai3
wait hold on
lilai3
  • lilai3
oh never mind so now you just basically solve for the delta t right and the answer would be in celsius?
ganeshie8
  • ganeshie8
what do you get for \(\Delta t\) ?
lilai3
  • lilai3
3.72 Celsius...
ganeshie8
  • ganeshie8
good \(\Delta t = 3.72 \) and you know that final temperature = \(39\) so initial temperature = ?
lilai3
  • lilai3
so subtract?
ganeshie8
  • ganeshie8
careful, water has lost some heat that means the initial temperature is greater than the final temperature
lilai3
  • lilai3
oh oops i read it wrong i mean add because the final temperature was the result of the two temperatures i accidentally interpreted it wrong 42.72 Celsius?
ganeshie8
  • ganeshie8
Yep, 42.72 C looks good
lilai3
  • lilai3
The fuel value of peanuts is 25 kj/ gram. If an average adult needs 2800 kilocalories of energy a day, what mass of peanuts would meet an average adult's energy needs for the day? Assume all of the fuel value of the peanuts can be converted to useful energy.
lilai3
  • lilai3
I am confused about what calories are.
ganeshie8
  • ganeshie8
dont get confused, all the problems you have been working on are similar... if we don't know what calorie means, we can always google :)
lilai3
  • lilai3
the main reason i suck at these problems is that my teacher just gives it to me along with the specific heat formula, so I don't get almost everything
ganeshie8
  • ganeshie8
Yes, that is all we ever need to work these problems First things first, lets put down the given info
lilai3
  • lilai3
25 kj/ grams 2800 kilocalories of energy a day we are finding mass
ganeshie8
  • ganeshie8
google gave me 1 kilocalorie = 4184 joules
lilai3
  • lilai3
yeah i think a calorie is like 4.18 something per joules.
ganeshie8
  • ganeshie8
right, so 2800 kilocalories = how many kilojoules ?

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