chemistry

I'm not sure how to solve this:

If 125 cal of heat is applied to a 60.0-g piece of copper at 20.0∘C , what will the final temperature be? The specific heat of copper is 0.0920 cal/(g⋅∘C). Express the final temperature numerically in degrees Celsius

Thanks

  1. 👍 0
  2. 👎 0
  3. 👁 2,925
  1. heatgained= mass*c*(Tf-Ti)

    Tf= 125/(60*.0920) + 20

    1. 👍 0
    2. 👎 0
  2. Good to hear from you again, MC. How is sister doing?

    1. 👍 0
    2. 👎 0
  3. I'm sorry, I haven't really understood what you've done. could you maybe break it down for me? I'm new at this whole chem thing. What is Tf and Ti?
    She's good, she's about to get her degree in education!

    Thanks
    -MC

    1. 👍 0
    2. 👎 0
  4. Tf is the final temperature, Ti is the initial temperature.

    1. 👍 0
    2. 👎 0
  5. I started from heat=mc deltaTemp
    then solved for final temperature Tf

    1. 👍 0
    2. 👎 0
  6. So what I'm getting as I set up the equation is 60g x 125 x 60
    and that doesn't seem right

    1. 👍 0
    2. 👎 0
  7. Hmmmm. that is far from right.

    Heatgained= mass*specificheat*change in temp

    change in temp= heatgaind/(specificheat*mass)

    change in temp= 125cal/(.0920cal/gC * 60g) = 33.6 check that

    chaknge in temp= finaltemp-initial temp

    so final temp= 33.6+20 C

    1. 👍 1
    2. 👎 0
  8. 53.6 c

    1. 👍 0
    2. 👎 1
    posted by James
  9. 47.6

    1. 👍 1
    2. 👎 0
    posted by Einstein

Respond to this Question

First Name

Your Response

Similar Questions

  1. chemistry

    If 125 cal of heat is applied to a 60.0-g piece of copper at 23.0∘C , what will the final temperature be? The specific heat of copper is 0.0920 cal/(g⋅∘C) .

    asked by gabriel on September 15, 2014
  2. Science

    A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000

    asked by Anonymous on September 26, 2011
  3. physics

    A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000

    asked by Allison on September 26, 2011
  4. Physics

    A substance has the following properties; Boiling Point = 140 °C, Freezing Point = -10 °C, specific heat as a gas, liquid and solid, 0.3 cal/g.C°, 1.8 cal/g.C° and 0.8 cal/g.C° respectively. Heat of fusion =100 cal/g and heat

    asked by Clarity on November 27, 2016
  5. Science -- Check my Answer Plz

    How much heat (calories) must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water? ** A) 540 cal ** B) 540 * 10^4 cal C) 800 cal D) 8000 cal Is A correct? or cud it be B.?

    asked by Saisir le Jour on January 4, 2016
  6. Chemistry

    What quantity of heat is necessary to convert 50.0 g of ice at 0.0 C into steam 100,0 C? The heat of fusion is 80.0 cal/g, the heat of vaporization is 540 cal/g, and the specific heat of water is 1.00 cal/gC.

    asked by Jill on March 27, 2012
  7. Chemistry/Physics

    How much heat is required to raise the temperature of 100 g of lead 20 degrees C? The specific heat of lead is 0.0305 cal/g K. A) 51 cal B) 61 cal C) 71 cal D) 81 cal

    asked by Domino on July 13, 2015
  8. science

    an aluminium container of mass 100g contains 200 g of ice at -20'c heat is added to the system at the rate of 100 cal/sec. what will be the final tmperature of the mixture after 4 min? given:specific heat of ice:0.5 cal/gm'c

    asked by anu on March 17, 2011
  9. Integrated Physics and Chemistry

    The specific heat capacity of copper is 0.09 cal/g°C. How much energy is needed to flow into a 10-gram sample to change its temperature from 20°C to 21°C? A)0.009 cal B)0.09 cal C)0.9 cal D)9 cal

    asked by Ashley!! on April 21, 2011
  10. Physics

    A 150 gram golden (specific heat 0.03 cal/gram^0 C) spoon at 20^0C is placed in a cup of tea at 100^0C. The amount of heat which the spoon absorbs from the tea which is necessary to reach spoon's tempature 47^0C is a)360.4 Cal

    asked by Vashti on March 5, 2015

More Similar Questions