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? 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

Question

Chemistry

Braydon Moss

5 September, 22:26

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? 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

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Answers (1)

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Rylie Bauer · Answer 1

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.