A 3.34 g sample of aluminum pellets (specific heat capacity = 0.89 J °C-1 g-1) and a 10.00 g sample of iron pe

Question

A 3.34 g sample of aluminum pellets (specific heat capacity = 0.89 J °C-1 g-1) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J °C-1 g-1) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 95.2 g of water at 21.7°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Answer 1

Please always try ALGEBRA!
Let the final temperature to be T°C. Notice that the heat capacity of liquid water is 4.182 J °C-1 g-1. We clearly have:
(3.34*0.89 + 10.00*0.45)*(100.0 - T) = 95.2*4.182*(T - 21.7)
or: 7.47*(100.0 - T) = 398*(T - 21.7)
Thus: T = (747+398*21.7)/(398+7.47) = 23.1
The final temperature of the metal and water mixture is 23.1°C.