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erma4kov [3.2K]
3 years ago
7

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
Chemistry
1 answer:
max2010maxim [7]3 years ago
6 0

Answer:

The answer to your question is: Initial temperature of copper = 67.1°C

Explanation:

Data

mass Copper = 248 g

volume Water = 390 ml

T1 water = 22.6°C

T2           = 39.9°C

T1 copper = ?

Specific heat water = 1 cal/g°C

Specific heat copper = 0.092 cal/g°C

Formula       copper             water

Heat is negative for copper because it releases heat

                  - mCp(T2 - T1) = mCp(T2 - T1)                  

                  - (248)(39.9 - T1) = 390 (1)((39.9 - 22.6)           Substitution

                 -9895.2 + 248T1 = 390(17.3)                             Simplification

                 -9895.2 + 248T1 = 6747

                 248 T1 = 6747 + 9895.2

                 248 T1 = 16642.2

                 T1 = 16642.2 / 248

                 T1 = 67.1 °C                                                         Result

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To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

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Thus, by combining them, we obtain:

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Which is related to the general line equation:

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Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

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