Question

1 Review | Constants | Periodic Table A 26.0 - g aluminum block is warmed to 65.1 °C and plunged into an insulated beaker containing 55.4 g water initially at 22.4 °C. The aluminum and the water are allowed to come to thermal equilibrium. Part A Assuming that no heat is lost, what is the final temperature of the water and aluminum? ALQ * 0 2 ? T = Submit Request Answer

Answer 1

In generally agained = alost we know that heat energy (Q) = M.CAT Hot compound will loss the heat Heat lost = Q = m. C; (T, -T) 1. To is not compound Initial temperature. cool compound (low Temperature) will gain the heat. Heat gain= q = ma că (T-Tp) 3 is cool compound Initial temperature. .cz (7- Ta) = m C, C1,- T) from the data 'Al' at high temperature, so it will lost the heat "Q", m, a mass = 269, C;= heat capacity -0 3/4. 2 Ti = 65.1°C

water will gain the heat & m₂ = 55.49, Ta= 22.4°C = heat capacity of water = 4.186 3/9.0 - Q, EQ m. c. (7,-T) = mg (T-I) 269. (6-902 m ) (65.1°C -1) = 55.48.(4.186.5.2 (-2242) 8 23: 452 g (65:12 - 1) = 234.9 €(T-22:40) - 65.19-T = 9.9 (T-22:49) => 65.1°C - T = 9.97 - 221.76 °C → 221.76 °C +65.1°C - 941 + 5 286.86 °C = 10.97 » T: 286.86 °C 10.9 = 26.3°C - 06 Final temperature of water and Al =) T=26.3°C