Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
The answer should be c because the relationship of pressure and plus the temperature decrease in the center.
