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Apr 7, 2016 at 6:33 vote accept Daniel Munoz
Apr 7, 2016 at 6:27 comment added Daniel Munoz Wait! like this? $\frac{k_{Al}L_{Pb}}{L_{Al}k_{Pb}} \Delta T_{Al} =\Delta T_{Pb} $ Then sub into the sum of $\Delta T_{overall}$
Apr 7, 2016 at 6:18 comment added Daniel Munoz So if I were to separate the temps as $T_1, T_2 and T_3$, $\Delta T_{overall}$ would equal $T_1-T_3$ right?
Apr 7, 2016 at 6:02 history answered John Rennie CC BY-SA 3.0