Question #c1c11
2 Answers
Explanation:
We can use the formula
The zinc has a greater heat content than the water, so when the zinc is placed into the water, it will lose heat (i.e., its temperature will decrease) and the water will gain heat (its temperature will increase). The final temperature should thus be somewhere in between
The final temperature of both will be the same, so the heat lost by the zinc will equal the heat gained by the water.
#-q_"Zn" = q_("H"_2"O")#
#-(m * c * DeltaT)_("Zn") = (m * c * DeltaT)_("H"_2"O")#
You can look up the specific heats of the zinc and water.
#-("17800 g" * ("0.39 J")/("g" * °"C") * (T_f - 67)°"C")_("Zn") = ("18000 g" * ("4.18 J")/("g" * °"C") * (T_f - 12.5)°"C")_("H"_2"O")#
Now the units cancel out.
#-("17800" cancel"g" * ("0.39 J")/(cancel"g" * cancel(°"C")) * (T_f - 67)cancel(°"C")) = "18000" cancel"g" * ("4.18 J")/(cancel"g" * cancel(°"C")) * (T_f - 12.5)cancel(°"C")#
#-("6942 J" * (T_f - 67)) = "75240 J" * (T_f - 12.5)#
#-("6942 T"_f - 465114 ) = "75240 T"_f - 940500#
#-"6942 T"_f + 465114 = "75240 T"_f - 940500#
#-"82182 T"_f = -1405614#
#T_f = 17.1 ^@"C"#
The final temperature of the water is 17.1 °C.
Explanation:
The guiding principle is the Law of Conservation of Energy: the sum of all the energy changes must add up to zero.
The formula for the heat
#color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "#
where
In this problem, there are two heat flows.
The final temperature of the zinc will be the same as the final temperature of the water.
In this problem,