Question

If you mix 30 mL of cold water with 70 mL of hot water in a calorimeter, then calculate that the cold water gained 142 J of heat and the hot water lost 181 J of heat, and the temperature change of the cold water (and calorimeter) was an increase in 1.93°C, then what is the heat capacity of the calorimeter in J/°C (only enter the number, not units, and assume that no heat was lost to the environment around the calorimeter, assume the density of water to be 1.00g/mL and specific heat capacity of water to be 4.184 J/g-°C)?

Answer 1

First, we need to calculate the heat gained by the cold water and the heat lost by the hot water:

Qcold = mcΔT = (30 g)(4.184 J/g-°C)(1.93°C) = 242.06 J

Qhot = mcΔT = (70 g)(4.184 J/g-°C)(-1.93°C) = -546.53 J

Since energy is conserved, we can assume that the heat gained by the cold water and calorimeter is equal to the heat lost by the hot water:

Qcold + Qcalorimeter = Qhot

Qcalorimeter = Qhot - Qcold

Qcalorimeter = -546.53 J - 242.06 J = -788.59 J

Therefore, the heat capacity of the calorimeter can be calculated as:

Ccalorimeter = Qcalorimeter / ΔT

Ccalorimeter = (-788.59 J) / (1.93°C)

Ccalorimeter ≈ -408.4 J/°C

Note that the negative sign indicates that the calorimeter loses heat when the system gains heat, which is expected since the calorimeter is absorbing some of the heat from the hot water.