Question

. A certain mass (m) of ice, initially at-5.00°C, is placed into a perfectly insulated container along with a certain mass (n) of steam (initially at 120°C). At equilibrium, there is only liquid water in the container. Write one completely detailed calorimetry equation necessary to solve for the equilibrium temperature of the water. Use symbols only (variables and constant names-no numbers), with all symbols defined (including Q's, AT's, etc.). You do not have to solve this equation or actually calculate the equilibrium temperature.

Answer 1

T be equilibrium temp ice m_1 at - 5 degree c steam m_2 at 120 degree c Q_1 = 334 J/g = heat of fusion of water Q_2 = 2.09 J/g degree c = heat of vaporization of water s_2 = specific heat of water = 4.18 J/g degree c s_3 = specific heat of steam = 2.09 J/g degree c delta T = 100 - T 20 rm_2 times s_3 + m_2 times Q_2 + m_2 times delta T times s_2 = 5m_1 times s_1 + m_1 Q_1 + m_1 s_1 times (100 - delta T)