An engineer wishes to determine the specific heat of a new metal alloy. A `0.150`-`kg` sample of the alloy is heated to `540°C`. It is then quickly placed in `400\ g` of water at `10°C`, which is contained a `200`-`g` aluminum calorimeter cup. The final temperature of the system is `30.5°C`. Calculate the specific heat of the alloy. Given: `T_a = 540°C` `m_a = 0.150\ kg` `T_w = 10°C` `m_w = 400\ g = 0.400\ kg` `T_(cal) = 10°C` `m_(cal) = 200\ g = 0.200\ kg` `T_f = 30.5°C` Known: `c_w = 4186\ J//kg•C°` `c_(cal) = 900\ J//kg•C°` Equation: `Q = mc∆T` heat lost = heat gained Solution: heat lost by alloy = heat gained by water + heat gained by calorimeter cup `m_aC_a∆T_a = m_wc_w∆T_w + m_(cal)c_(cal)∆T_(cal)` `c_a = (m_wc_w∆T_w + m_(cal)c_(cal)∆T_(cal))/(m_a∆T_a)` `c_a = ((0.400\ kg)(4186\ J//kg•C°)(30.5°C-10°C) + (0.200\ kg)(900\ J//kg•C°)(30.5°C-10°C))/((0.150\ kg)(540°C-30.5°C))`