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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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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.

Calorimetry is the measurement technique, in which heat of the chemical reactions and heat capacity is measured. The value of Ccal for the calorimeter is 99 J/K.

According to the first law of thermodynamics, the energy cannot be created nor be destroyed. It can be transferred from one form to another.

Such as:

  • Heat released by the the hot water at 345K  = Heat absorbed by the cold water at 298 K + Heat absorbed by the calorimeter

Also, from the formula, it can be calculated as:

  • Heat or H = mass × Specific heat × change in temperature

Given that,

  • Density of water = 0.997 g/ ml
  • Mass of water = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g
  • Specific heat = 1 cal/ g Celcius

Now, substituting the values, heat released by the water will be:

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

Similarly, heat absorbed by the cold water will be:

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

Now, the overall heat absorbed by the calorimeter will be equal to:

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

The value of Ccal will be:

  • 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 or 98.6 J/K (in Joules)

Therefore, the Ccal for the calorimeter will be equal to approximately 99 J/K.

To know more about calorimeter, refer to the following link:

https://brainly.com/question/15187880?referrer=searchResults

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