Astudent 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

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:

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

Given that,

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

Similarly, heat absorbed by the cold water will be:

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

The value of Ccal will be:

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

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

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Explanation:

1) First law of thermodynamic (energy balance)

2) Energy change of each substance:

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

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

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

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

4) Balance

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

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

       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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%