A 15.1 g aluminum block is warmed to 53.2 °C and plunged into an insulated beakercontaining 32.6 g of water initially at 24.4 °C. The aluminum and the water are allowed to cometo thermal equilibrium.Assuming that no heat is lost, what is the final temperature of the water and aluminum?To solve this problem, you need the heat capacity of water and Al, I know off hand the heat capacity of water but you\'ll need to look it up for Al. From the problem, you know that the heat given up as the aluminum cools is gained by the water heating up. As there is no phase change, the heat gained (or lost) is mass * heat capacity * temperature change energy gained by water = energy lost by Al. Let Tf be the final temperature of the Al and water. 32.6 g * 4.18 J/gK * (Tf - 24.4C) = 15.1 g * heat capacity of Al * (53.2 C - Tf) Solve for Tf. Because there is almost twice as much water as Al and water has a higher heat capacity than Al, the temperature will be closer to 24.4C than it is to 53.2COther related question