Question

cal) for her coffee cup calorimeter. She pours a 50.0 mL sample of water at 345 K into the calorimeter containing a 50.0 mL sample of water at 298 K. She carefully records the final temperature of the water as 317 K. What is the value of Ccal for the calorimeter? A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (Ccal) for her coffee cup calorimeter. She pours a 50.0 mL sample of water at 345 K into the calorimeter containing a 50.0 mL sample of water at 298 K. She carefully records the final temperature of the water as 317 K. What is the value of Ccal for the calorimeter? 99 J/K 21 J/K 76 J/K 28 J/K 19 J/K

Answer 1

Answer:

• First choice: 99 J/K

Explanation:

1) First law of thermodynamic (energy balance)

• Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

2) Energy change of each substance:

• General formula:

     

Heat released or absorbed = mass × Specific heat × change in temperature

• density of water: you may take 0.997 g/ ml as an average density for the water.

• mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

• Specif heat of water: 1 cal / g°C

• Heat released by the hot water:

       Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

• Heat absorbed by the cold water:

       Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

• Heat absorbed by the calorimeter

       Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

4) Balance

• Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

• Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

• Convert to cal / K to Joule / K

• 1 cal = 4.18 Joule

       23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.