Question

A 7.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 14.00-g sample of iron pellets (specific heat capacity = 0.45 J/°C·g) are heated to 100.0 °C. The mixture of hot iron and aluminum is then dropped into 71.3 g water at 22.0 °C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Answer 1

heat released by both the metal mixture will be equal to heat gained by water

m(Fe)*S(Fe)*(Ti-Tf) + m(Al)*S(Al)*(Ti-Tf) = m(H2O)*S(H2O)*(Tf-Ti)

so 14.00*0.45*(100-x) + 7.00*0.89*(100-x) = 92.3*4.184*(x-22)

x= 24.45 degree centigrade

so final temperature is 24.45^oC