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2) The
"tool" I provide you for solving this kind of question is: "Whenever two
things with different s.h.c.'s, masses, and initial temeratures, it is
alway true that the heat gained by the one at lower temperature is the
same heat lost by the one at higher temperature". In this question,
the heat lost = 1 cal/g°C x 10
g x (Tf – 50 °C) since
we don't know what the final temperature is and just give it as Tf;
and the heat gained = 1 cal/g °C x
5 g x (Tf – 20 °C).
If you consider heat gained as "+", and heat lost as "–" (like when you
balance you check as we discussed in the class), you can get the following
equation:
–[1 cal/g °C x 10 g x (Tf
– 50 °C)] = 1 cal/g °C x 5
g x (Tf – 20 °C)
Now, solve Tf for the above equation!
Knowing this,
you can basically solve any question related to this using plugging in
different s.h.c.'s, masses, and initial temeratures, then solve the final
temperature. Now, work on the question: "When you mix 10.0
g of iron (s.h.c. = 0.11 cal/g °C) at 10 °C with 2.0 g of water
(s.h.c. = 1.0 cal/g °C) at 60 °C, the final temperature?"
The equation you need to solve is now:
–[1.0 cal/g °C x 2 g x
(Tf – 60 °C)] = 0.11 cal/g °C x
10.0 g x (Tf – 10 °C)
(You may now solve this equation!)