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  1. #1
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    Exclamation Reps for chemistry question help!!!!! (srs)

    How much heat energy is required to raise the temperature of 0.359 kg of copper from 23.0 C to 60.0 C If the specific heat of copper is 0.0920 cal/(g x C)


    That is a times sign in parenthesis and C stands for degrees celsius, I cant figure out how to do superscript on here.
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  2. #2
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    bout 350
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  3. #3
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    Q = mcΔT

    Q is heat, just plug in and solve.

    also convert your kg's to g
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  4. #4
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    13.3 calories
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  5. #5
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    Originally Posted by CurlsfotheGirls View Post
    is this srs?

    do you have no formulas?

    even still you should be able to solve this.
    I am horrible at chemistry man, retaking this course AGAIN

    formula is SH= Heat
    ___________
    Mass x Delta T (temperature change)
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  6. #6
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    Originally Posted by UnflappedAcorn View Post
    13.3 calories
    not going through brah.
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  7. #7
    Proudest manlet Shreder54's Avatar
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    1222? i forget the damn formula
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  8. #8
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    Originally Posted by CurlsfotheGirls View Post
    this

    and check your units.

    if you need to; make sure to convert from cal to Joules.

    edit: if your not gettting it post brah and i got you
    so the Kg to G is 359, Joules is fuaaaaaaaaaaaaark!!!!!!1 I cant get this **** man.
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  9. #9
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    Originally Posted by Shreder54 View Post
    1.22? i forget the damn formula
    Forgot to convert kilograms to grams
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  10. #10
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    Originally Posted by lift_heavy20 View Post
    not going through brah.
    maybe its the significant units

    the exact is: 13.283
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  11. #11
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    Originally Posted by lift_heavy20 View Post
    so the Kg to G is 359, Joules is fuaaaaaaaaaaaaark!!!!!!1 I cant get this **** man.
    1 cal = 4.186 J I think

    I haven't taken chem in like 3 years, this is stuff I remember from physics
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  12. #12
    Proudest manlet Shreder54's Avatar
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    Originally Posted by captiontom View Post
    Forgot to convert kilograms to grams
    Yeah you're right, i thought I did. It's 1222 then I think.

    It's really easy if you look in book or look it up online...
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  13. #13
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    Q = mcΔT
    Q = 359g*0.0920*37
    = 1222.036 cal

    1 cal = 4.184 J
    1222.036cal = 5112.998J
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  14. #14
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    thanks guys it was actually 1222, anyone want to explain how this was achieved?
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  15. #15
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    I think I repped everyone in here thanks guys, plz explain how answer was achieved?
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  16. #16
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    Originally Posted by lift_heavy20 View Post
    I think I repped everyone in here thanks guys, plz explain how answer was achieved?

    there is a formula...q=mc(delta)T

    m is the mass of the copper
    c is the specific heat
    deltaT is the change in temp (in kelvin= C + 273)
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  17. #17
    Registered User fecalneek's Avatar
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    Originally Posted by lift_heavy20 View Post
    thanks guys it was actually 1222, anyone want to explain how this was achieved?
    you don't need to do any cal to J conversions.

    just convert your kg to g so you get 359g.

    Q = mcΔt

    m=359 g
    c is the specific heat, in this case it's copper
    Δt is the change in temperature so it's 60-23

    just plug that all in and you should get 1222
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  18. #18
    Registered User captiontom's Avatar
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    Originally Posted by lift_heavy20 View Post
    thanks guys it was actually 1222, anyone want to explain how this was achieved?
    First convert kilograms of copper to grams

    Second determine the change in temperature

    Third multiply the specific heat by the grams of copper by the change in temperature

    The answer is in calories
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  19. #19
    Registered User thercias's Avatar
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    it's just a basic formula equation bro, don't you know the formula?

    q= mcT
    = 35.9g*0.092*37C
    = ....

    had an chem exam this monday and this was one of the chapters
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  20. #20
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    Originally Posted by thercias View Post
    it's just a basic formula equation bro, don't you know the formula?

    q= mcT
    = 35.9g*0.092*37C
    = ....

    had an chem exam this monday and this was one of the chapters
    And you missed this problem...
    359 grams not 35.9
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  21. #21
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    I love working with copper at work * plumber *
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  22. #22
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    Originally Posted by Leh View Post
    Q = mcΔT
    Q = 359g*0.0920*37
    = 1222.036 cal

    1 cal = 4.184 J
    1222.036cal = 5112.998J


    somebody repp this mawfukkah right now


    EDIT: Strong first post brah
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