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lilai3

  • one year ago

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?

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  1. ganeshie8
    • one year ago
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    so the temperature of the water has risen by \(45-22 = 23^{\circ}C\)

  2. lilai3
    • one year ago
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    that's the delta t right?

  3. ganeshie8
    • one year ago
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    Yes, then the heat added to the water is given by, \(Q=c*m*\Delta t=4.18*42800*23 = 4114792 J\)

  4. ganeshie8
    • one year ago
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    That means the bar has lost the same amount of heat. You can use that to find the mass of bar

  5. lilai3
    • one year ago
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    Oh I see.

  6. lilai3
    • one year ago
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    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?

  7. ganeshie8
    • one year ago
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    0.446 is specific heat of iron is it ?

  8. lilai3
    • one year ago
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    yeah

  9. ganeshie8
    • one year ago
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    Looks good !

  10. lilai3
    • one year ago
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    thanks can you also help me with this one?

  11. lilai3
    • one year ago
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    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?

  12. ganeshie8
    • one year ago
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    looks like a similar problem.. give it a try

  13. lilai3
    • one year ago
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    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

  14. ganeshie8
    • one year ago
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    First, write down the given info Water : \(m = 752 cm^3 = 752 g\) \(c = 4.18 J/G^{\circ}C\) \(\Delta t = ?\) Aluminium pan : \(m = 1kg= 1000 g\) \(c = 0.9 J/G^{\circ}C\) \(\Delta t = 13\)

  15. lilai3
    • one year ago
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    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)?

  16. ganeshie8
    • one year ago
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    Energy released by water = Energy absorbed by aluminium pan

  17. ganeshie8
    • one year ago
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    \[4.18*752*\Delta t = 0.9*1000*13 \] you can solve \(\Delta t\)

  18. lilai3
    • one year ago
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    oh okay and would that be 39-26?

  19. lilai3
    • one year ago
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    wait hold on

  20. lilai3
    • one year ago
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    oh never mind so now you just basically solve for the delta t right and the answer would be in celsius?

  21. ganeshie8
    • one year ago
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    what do you get for \(\Delta t\) ?

  22. lilai3
    • one year ago
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    3.72 Celsius...

  23. ganeshie8
    • one year ago
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    good \(\Delta t = 3.72 \) and you know that final temperature = \(39\) so initial temperature = ?

  24. lilai3
    • one year ago
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    so subtract?

  25. ganeshie8
    • one year ago
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    careful, water has lost some heat that means the initial temperature is greater than the final temperature

  26. lilai3
    • one year ago
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    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?

  27. ganeshie8
    • one year ago
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    Yep, 42.72 C looks good

  28. lilai3
    • one year ago
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    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.

  29. lilai3
    • one year ago
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    I am confused about what calories are.

  30. ganeshie8
    • one year ago
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    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 :)

  31. lilai3
    • one year ago
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    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

  32. ganeshie8
    • one year ago
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    Yes, that is all we ever need to work these problems First things first, lets put down the given info

  33. lilai3
    • one year ago
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    25 kj/ grams 2800 kilocalories of energy a day we are finding mass

  34. ganeshie8
    • one year ago
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    google gave me 1 kilocalorie = 4184 joules

  35. lilai3
    • one year ago
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    yeah i think a calorie is like 4.18 something per joules.

  36. ganeshie8
    • one year ago
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    right, so 2800 kilocalories = how many kilojoules ?

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