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melisa1
27 days ago
7

An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroun

dings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?
Chemistry
1 answer:
Anarel [2.6K]27 days ago
5 0

Answer:

The final temperature of the entire water mixture, once all the ice has melted, stands at 12.9°C. It’s crucial to understand that in a closed system where no heat is lost, the total heat exchanged is 0.

This implies that the temperature decreased as the ice transferred heat to the water, leading to its cooling.

Explanation:

To begin:

Q1 = Q representing the heat gained from melting ice.

Q2 = Q denoting the heat lost by the water to melt the ice.

Q1 + Q2 = 0

Given that the ice starts at 0 °C, we must first determine the energy required to completely melt it. If the ice had been at a lower temperature, we would have raised it to 0 °C using the formula:

Q = mass × specific heat × (ΔT)

and then, to perform the state transition by utilizing the latent heat of fusion.

The heat of fusion for water at 0 °C is roughly 334 joules per gram.

Thus, Q = Hf × mass.

Q1 = 334 J/g × 8.32 g = 2778.88 J

For the water, we use:

Q = mass × specific heat × (ΔT)

Q2 = 55g × 4180 J/kg·K (Tfinal - T initial)

Converting 55 g to kg yields 0.055kg to maintain consistent units.

Q2 = 0.055kg × 4180 J/kg·K (Tfinal (unknown) - 25°)

Note that the temperature in degrees Kelvin (K) is the same for specific heat but differentiates as K vs. °C.

25°C = 298K

Q2 = 0.055kg × 4180 J/kg·K (Tfinal - 298K)

The conclusion:

Q1 + Q2 = 0

334 J/g × 8.32 g + 0.055kg × 4180 J/kg·K (Tfinal - 298K)

2778.88 J + 229.9 J/K (Tfinal - 298 K) = 0

2778.88 J + 229.9 J/K × Tfinal - 68510.2 J = 0

229.9 J/K × Tfinal = 68510.2 J - 2778.88 J

Tfinal = 65731.4 J / 229.9 K/J

Tfinal = 285.9 K

Tfinal = 285.9 K - 273K = 12.9 °C

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Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K

Starting temperature of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

mC \Delta T = mC \times \Delta T

10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

T_{f} - 100^{o}C = 0^{o}C - T_{f}

2T_{f} = 100^{o}C

T_{f} = 50^{o}C

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

\Delta H = mC \Delta T

= 10000 g \times 0.385 J/K g \times 323 K

= 1243550 J

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Next, determine the entropy change for block 1 as follows.

\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

= 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

= 10000 g \times 0.385 J/K g \times -0.143

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

                   = -554.12 J/K + 647.49 J/K\Delta S_{total} = \Delta S_{1} + \Delta S_{2}

           = 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr
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Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 × 10^{-5}

Calculating pKb: pKb = - log (6.3 × 10^{-5})

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

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Verification: The typical range for pKa in weak acids is from 2 to 13.

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A tritium nucleus is formed by combining two neutrons and a proton. the mass of this nucleus is 9.106 × 10–3 universal mass unit
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1 month ago
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Answer :

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Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

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Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

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