Question

Calculate the calorimeter constant with the following

first water 50 ml

first temp. 21.5 C

add more substance another 50 ml

temp. of substance added 60 C

Final temp 39.6 C

Answer 1

The substance at 60 ^oC will lose heat, where as water at 21.5 ^oC and calorimeter will gain heat.

Added substance:

Final temeprature, Tf = 39.6 ^oC

Initial temperature of the substance added, Ti = 60 ^oC

Mass of substance at 60 ^oC, = 50 mL * (1.00 g / mL) = 50 g

=> Total heat lost by the substance, Qt = m*s*T = m*s*(Tf - Ti)

=> Qt = 50 g * 4.184 J/g.^oC * (39.6 - 60) ^oC = - 4267.68 J

First water:

Mass of water at  21.5 ^oC, = 50 mL * (1.00 g / mL) = 50 g

=> Heat gained by water at 21.5 ^oC(Ti = 21.5 ^oC), Qw = m*s*T = m*s*(Tf - Ti)

=> Qw = 50 g * 4.184 J/g.^oC * (39.6 - 21.5) ^oC = 3786.52 J

Calorimeter:

Initial temperature of calorimeter, Ti = 21.5 ^oC

Final temperature of calorimeter, Tf = 39.6 ^oC

=> Heat gained by calorimeter, Qc = C * T = C * (39.6 - 21.5) = 18.1 C J

where "C J/^oC " is heat capacity of calorimeter

Total heat gained = Qw + Qc =  3786.52 J + 18.1 C J

According to law of calorimetry

- (Total heat lost, Qt) = total heat gained (Qw + Qc)

=> - ( - 4267.68) J = 3786.52 J + 18.1 C J

=> 18.1 C = 4267.68 - 3786.52 = 481.16

=> C =  481.16 / 18.1

=> C =  26.6 J/^oC (Answer)

Hence calorimeter constant = 26.6 J/^oC