Asked by jean luc
If a hot piece of metal (at 80 degrees Celsius) is placed in cold water (10 degrees Celsius), what can you say about the final temperature of the water/metal mixture? If you wanted a more accurate answer, what other information would you need?
Answers
Answered by
DrBob222
I'm sure you know what will happen so I'll leave the first part to you. To be more quantitative, you need the following:
q(heat lost by metal) = mass metal x specific heat metal x (Tfinal-Tinitial).
q(heat gained by water) = mass H2O x specific heat water x (Tfinal-Tinitial).
You can go through the two equations and count up what other information you would need.
q(heat lost by metal) = mass metal x specific heat metal x (Tfinal-Tinitial).
q(heat gained by water) = mass H2O x specific heat water x (Tfinal-Tinitial).
You can go through the two equations and count up what other information you would need.
Answered by
jean luc
Oooooooooooh....thanks Dr. Bob cannot believe I missed that
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