If `200\ cm^3` of tea at `95°C` is poured into a `150-g` glass cup initially at `25°C`, what will be the common final temperature `T` of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings? Given: `V_(tea) = 200\ cm^3 = 2.00 times 10^-4\ m^3` `T_(tea) = 95°C` `m_(c u p) = 150\ g = 0.150\ kg` Known: `c_(tea) = 4186\ J//kg•C°` `c_(cu p) = 840\ J//kg•C°` `ρ_(tea) = 1.0 times 10^3\ kg//m^3` Equaiton: `Q = mc∆T` `m = ρV` Solution: `m_(tea) = ρ_(tea)V_(tea)` `m_(tea) = (1.0 times 10^3\ kg//m^3)(2.00 times 10^-4\ m^3)` `m_(tea) = 0.20\ kg` heat lost by tea = heat gained by cup `m_(tea)c_(tea)(95°C - T) = m_(cu p)c_(cu p)(T - 25°C)` `(0.20\ kg)(4186\ J//kg•C°)(95°C - T) = (0.150\ kg)(840\ J//kg•C°)(T - 25°C)` `79500\ J - (837\ J//C°)T = (126\ J//C°)T - 3150\ J`