A 23.35 g aluminum block is warmed to 60.35 °C and plugged into an insulated beaker containing 90.93 g water initially at 19.81 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?
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heat lost by Al = heat gained by water
deltaT Al = starting temp - final temp
deltaT H2O = final temp - starting temp
23.35g Al x 0.900J/g-C x (60.35 - Tf) = 90.93g x 4.184J/g-C x (Tf - 19.81)
-21.915 Tf + 1368.255 = 380.451Tf - 7536.737
8904.499 = 402.366Tf
22.13C = Tf
deltaT Al = starting temp - final temp
deltaT H2O = final temp - starting temp
23.35g Al x 0.900J/g-C x (60.35 - Tf) = 90.93g x 4.184J/g-C x (Tf - 19.81)
-21.915 Tf + 1368.255 = 380.451Tf - 7536.737
8904.499 = 402.366Tf
22.13C = Tf