In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 3.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol. Assume that the specific heat of the solution form

First, we find the heat released by the dissolution of CaCl₂. The grams will be converted to moles using the molar mass (110.986 g/mol), then multiplied by the molar heat of solution:

(3.00 g)(mol/110.986 g)(-82.8 kJ/mol) = -2.23812 kJ

The negative sign indicates that heat is released. Extra significant figures are included to avoid round-off errors.

The amount of heat released by the CaCl₂ dissolution is equal to the heat absorbed by the water. The equation is rearranged to solve for Δt, the temperature change of the water.

Q = mcΔt ⇒ Δt = Q/(mc)

Δt = (2.23812 kJ)(1000 J/kJ) / (100 mL)(1 g/mL)(4.184 Jg⁻¹ °C⁻¹) = 5.3 °C

We can then calculated the final temperature t₂ of the solution:

Δt = t₂ - t₁

t₂ = Δt + t₁ = 5.3 °C + 23.0 °C = 28.3 °C

dsdrajlin dsdrajlin · Answer 2 · 2021-10-17T23:45:12+02:00

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

First, we will convert 3.00 g of CaCl₂ to moles using its molar mass (110.98 g/mol).

[tex]3.00 g \times \frac{1mol}{110.98g} = 0.0270 mol[/tex]

The heat of solution (ΔHsoln) of CaCl₂ is −82.8 kJ/mol. The heat released by the solution of 0.0270 moles is:

[tex]0.0270 mol \times \frac{-82.8kJ}{mol} = -2.24 kJ[/tex]

According to the law of conservation of energy, the sum of the heat released by the solution (Qs) and the heat absorbed by the calorimeter (Qc) is zero.

[tex]Qs + Qc = 0\\\\Qc = -Qs = 2.24 kJ[/tex]

Assuming the density of water is 1 g/mL, we have 100 mL (100 g) of water and 3.00 g of CaCl₂. The mass of the solution (m) is:

[tex]m = 100g + 3.00 g = 103 g[/tex]

Finally, we can calculate the final temperature of the system using the following expression.

[tex]Qc = c \times m \times (T_2 - T_1)[/tex]

where,

c: specific heat of the solution (same as water 4.18 J/g.°C)

T₁ and T₂: initial and final temperature

[tex]T_2 = \frac{Qc}{c \times m} + T_1 = \frac{2.24 \times 10^{3}J }{(\frac{4.18J}{g.\° C} ) \times 103 g} + 23.0 \° C = 28.2 \° C[/tex]

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

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