Goal Solve a calorimetry problem involving three substances at three different temperatures. Problem Suppose 0.400 kg of water initially at 40.0°C is poured into a 0.300-kg glass beaker having a temperature of 25.0°C. A 0.500-kg block of aluminum at 37.0°C is placed in the water and the system insulated. Calculate the final equilibrium temperature of the system. Strategy The energy transfer for the water, aluminum, and glass will be designated Qw, Qal, and Qg, respectively. The sum of these transfers must equal zero, by conservation of energy. Construct a table, assemble the three terms from the given data, and solve for the final equilibrium temperature, T. SOLUTION Apply the final temperature equation to the system. (1) Qw + Qal + Qg = 0 (2) mwcw(T -Tw) + malcal(T -Tal) + mgcg(T -Tg) = 0 Construct a data table. Q (J) m (kg) c (J/kg · °C) Tf Ti Qw 0.400 4190 T 40.0°C Qal 0.500 9.00 102 T 37.0°C Qg 0.300 837 T 25.0°C Using the table, substitute into Equation (2). (1.68 J cross 103 J/°C)(T - 40.0°C) + (4.50 102 J/°C)(T - 37.0°C) + (2.51 102 J/°C)(T - 25.0°C) = 0 (1.68 J cross 103 J/°C + 4.50 102 J/°C + 2.51 102 J/°C)T = 9.01 104 J T = 37.9°C Use the worked example above to help you solve this problem. Suppose 0.405 kg of water initially at 39.5°C is poured into a 0.300 kg glass beaker having a temperature of 25.0°C. A 0.500 kg block of aluminum at 37.0°C is placed in the water, and the system insulated. Calculate the final equilibrium temperature of the system. °C