Question

A metal sample weighinh 45.2 g and at a temperature of 100.0 C was placed in 38.6 g of water in an aluminum calorimeter at 25.2 C. The mass of the calorimeter is 70.4 g and its specific heat is 0.900 J/gC. At equilibrium the temperature of the water, metal and calorimeter was 33.0 C.

A. How much heat flowed into the water and calorimeter? Total heat gained = heat gained by calorimeter + heat gained by water

OR q=(mass)(s)(changeT) + (mass)(s)(changeT)

Answer 1

mass of water = 38.6 g

specific heat of water = 4.184 J/g.^oC

temperature change for water = final temperature - initial temperature

temperature change for water = (33.0 ^oC) - (25.2 ^oC)

temperature change for water = 7.8 ^oC

heat gained by water = (mass of water) * (specific heat of water) * (temperature change for water)

heat gained by water = (38.6 g) * (4.184 J/g.^oC) * (7.8 ^oC)

heat gained by water = 1259.72 J

mass of aluminum calorimeter = 70.4 g

specific of aluminum = 0.900 J/g.^oC

temperature change for aluminum calorimeter = final temperature - initial temperature

temperature change for aluminum calorimeter = (33.0 ^oC) - (25.2 ^oC)

temperature change for aluminum calorimeter = 7.8 ^oC

heat gained by aluminum calorimeter = (mass of aluminum) * (specific heat of aluminum) * (temperature change for water)

heat gained by aluminum calorimeter = (70.4 g) * (0.900 J/g.^oC) * (7.8 ^oC)

heat gained by aluminum calorimeter = 494.21 J

Total heat gained by water and calorimeter = 1259.72 J + 494.21 J

Total heat gained by water and calorimeter = 1753.93 J