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Chapter 15: Problem 69

Estimate the entropy change of \(850 \mathrm{g}\) of water when it is heated from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C} .\) [Hint: Assume that the heat flows into the water at an average temperature. \(]\)

Short Answer

Expert verified
The estimated entropy change is approximately 349 J/K when 850g of water is heated from 20.0掳C to 50.0掳C.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we will add 273.15 to each Celsius temperature: \(T_1 = 20.0 + 273.15 = 293.15 \mathrm{K}\) \(T_2 = 50.0 + 273.15 = 323.15 \mathrm{K}\)
02

Calculate the average temperature

We will now calculate the average temperature during the heating process: \(T_\text{avg} = \frac{T_1 + T_2}{2} = \frac{293.15 + 323.15}{2} = 308.15 \mathrm{K}\)
03

Calculate the entropy change

We can now calculate the entropy change using the formula: \(\Delta S = mC\log \frac{T_2}{T_1} = (850 \mathrm{g})(4.18 \mathrm{J / g \cdot K})\log \frac{323.15 \mathrm{K}}{293.15 \mathrm{K}}\) Now, compute the logarithm and multiply the resulting value with the mass and specific heat capacity: \(\Delta S = (850 \mathrm{g})(4.18 \mathrm{J / g \cdot K})(0.0970) = 348.803 \mathrm{J/K}\)
04

Round the result

Finally, we will round the result to an appropriate number of significant figures (three): \(\Delta S \approx 349 \mathrm{J/K}\) The estimated entropy change of 850g of water when it is heated from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) is approximately 349 J/K.

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Most popular questions from this chapter

How much heat does a heat pump with a coefficient of performance of 3.0 deliver when supplied with \(1.00 \mathrm{kJ}\) of electricity?
Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$
Two engines operate between the same two temperatures of \(750 \mathrm{K}\) and \(350 \mathrm{K},\) and have the same rate of heat input. One of the engines is a reversible engine with a power output of \(2.3 \times 10^{4} \mathrm{W} .\) The second engine has an efficiency of \(42 \% .\) What is the power output of the second engine?
List these in order of increasing entropy: (a) 1 mol of water at $20^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)20^{\circ} \mathrm{C}$ in separate containers; (b) a mixture of 1 mol of water at \(20^{\circ} \mathrm{C}\) and 1 mol of ethanol at \(20^{\circ} \mathrm{C} ;\) (c) 0.5 mol of water at \(20^{\circ} \mathrm{C}\) and 0.5 mol of ethanol at \(20^{\circ} \mathrm{C}\) in separate containers; (d) a mixture of 1 mol of water at $30^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)30^{\circ} \mathrm{C}$.
A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
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