Question

A 61.18 gram sample of iron (with a heat capacity of 0.450 J/g℃) is heated to 100.00。It is then transferred to a coffee cup calorimeter containing 52.33 g of water (specific heat of 4.184 J/ g℃) initially at 20.67 ℃. If the final temperature of the system is 28.40, what was the heat gained by the calorimeter? If the calorimeter had a mass of 27.88 g, what is the heat capacity of the calorimeter? J absorbed by the calorimeter

Answer 1

Heat absorbed/released is calculated using the formula:

Q = m*C*dT

m = mass, C = specific heat, dT = temperature change, Q = heat change

First we calculate heat absorbed by water, Qw.

Qw = m*C*dT = 52.33*4.184*(28.40-20.67) = 1692.47 J

Now, we calculate the heat lost by copper.

Q_Cu = m*C*dT = 61.18*0.45*(100-28.40) = 1971.22 J

Since heat lost by copper is more than that absorbed by water, so the remaining heat is absorbed by calorimeter.

Heat absorbed by calorimeter, Qc = Q_Cu - Qw = 1971.22-1692.47 = 278.75 J

Mass of calorimeter = 27.88 g

So, Heat capacity of calorimeter, C = Qc/(m*dT) = 278.75/(27.88*(28.40-20.67)) = 1.293 J/(g.C)