12.0 ounces of water are heated during the preparation of a cup of coffee. 1.138X10^3 J of heat are added to the water, which is initially at 19.7 degrees celsius. What's the final temp of the coffee?

asked by @stephanier47 • over 1 year ago • ChemistryIdentifying Acids and Bases • 5 pts

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1 answer

Stephanie, this is another thermochemistry problem. We have water being heated, with initial and final temperatures. I hope you realize that we need to use q=mcdeltaT here.

They want you to calculate the final temperature of coffee, which will be the same final temp for water. Q will be positive because the system is gaining heat.

  • +q= mcdeltaT
  • 1.13810^3 J= 340.2 g *(4.184J/(gdeg C))(Tfinal-19.7 deg C)
  • We start with 12 ounces of water, convert to grams by using 1 ounce=28.35 grams. Also, specific heat capacity of water is c=4.184 J/(gC), just remember that.
  • delta T= (Tfinal -19.7)= 0.7994 deg C
  • Tfinal=0.7994+19.7= 20.50 deg C

Hope that helped!

answered by @dashab2 • over 1 year ago