A calorimeter is to be calibrated: 72.55 g of water at 71.6 degree C added to a

Question

A calorimeter is to be calibrated: 72.55 g of water at 71.6 degree C added to a calorimeter containing 58.85 g of water at 22.4 degree C. After stirring and waiting for the system to equilibrate, the final temperature reached 47.3 degree C. Calculate the heat capacity of the calorimeter with appropriate units and significant figures. (The specific heat capacity of water is 4.184 J g^-1 degree C^-1) Energy lost by the hot water: q = m C_p Delta T Energy gained by the cold water: q = m C_p Delta T Heat capacity of the calorimeter

Explanation / Answer

Q1.

energy lost by water

Q = m*C*(Tf-Ti)

Q = 72.55*4.184*(47.3-71.6)

Q = -7376.24556 J lost

Q2.

energy gained by cold water

Q = m*C*(Tf-Ti)

Q = 58.85*4.184*(47.3-22.4)

Q = 6131.08716 J gained

Q3.

Qtotal = Qgain + Qlost + Qcalorimter

Since this is adiabatic, i.e. no loss, then

Q total = 0

Qgain + Qlost + Qcalorimter = 0

6131.08716 + Qcalorimter - 7376.24556 = 0

Qcalorimter = 7376.24556 - 6131.08716 = 1245.1584 J

Heat capacity for calorimter:

dT = 47.3-22.4 = 24.9 °C

so

C = q/dT = 1245.1584 /24.9 =50.0 J/°C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.