Using the data in Table 3.2, determine the difference between [latex]c_p[/latex] and [latex]c_v[/latex] for saturated liquid water at 20.0°C.

From Table 3.2, for water, we find that Table 3.2 Values of [latex]β[/latex] and [latex]κ[/latex] for Various Liquids at [latex]20.°\text{C} (68°\text{F})[/latex] Substance [latex]β × 10^6[/latex] [latex]κ × 10^11[/latex] [latex]\text{R}^{−1}[/latex] [latex]\text{K}^{−1}[/latex] [latex]\text{ft}^2/\text{lbf}[/latex] [latex]\text{m}^2/\text{N}[/latex] Benzene 0.689 1.24 4550 95 Diethyl ether 0.922 1.66 8950 187 Ethyl alcohol 0.622 1.12 5310 111 Glycerin 0.281 0.505 1010 21 Heptane (n) 0.683 1.23 6890 144 Mercury 0.101 0.182 192 4.02 Water 0.115 0.207 2200 45.9 Source: Adapted by permission of the publisher from Zemansky, M. W., Abbott, M. M., Van Ness, H. C., 1975. Basic Engineering Thermodynamics, second ed. McGraw Hill, New York. [latex]β = 0.207 × 10^{−6} \text{K}^{−1}[/latex] [latex]κ = 45.9 × 10^{−11} \text{m}^2/\text{N}[/latex] and, from Table C.1b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics , we find that [latex]v = v_f(20.0°\text{C}) = 0.001002 \text{m}^3\text{/kg}[/latex] Then, from Eq. (11.30), we have [latex]c_p − c_v = \frac{Tβ^2v}{κ} = \frac{(293 \text{K})(0.207 × 10^{−6} \text{K}^{−1})^2 (0.001002 \text{m}^3\text{/kg})}{45.9 × 10^{−11} \text{m.s}^2\text{/kg}} [/latex] [latex] = 2.74 × 10^{−5} \text{J/(kg.K)} = 2.74 × 10^{−8} \text{KJ/(kg.K)}[/latex] In most applications, this difference is clearly negligible, since the value of [latex]c_p[/latex] for liquid water at standard temperature and pressure is [latex]4.18 \text{KJ/(kg.K)}[/latex].

Question 11.10: Using the data in Table 3.2, determine the difference betwee...

Modern Engineering Thermodynamics

Using the data in Table 3.2, determine the difference between $c_p$ and $c_v$ for saturated liquid water at 20.0°C.

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From Table 3.2, for water, we find that

Table 3.2 Values of $β$ and $κ$ for Various Liquids at $20.°\text{C} (68°\text{F})$
Substance	$β \times 10^6$		$κ \times 10^11$
Substance	$\text{R}^{−1}$	$\text{K}^{−1}$	$\text{ft}^2/\text{lbf}$	$\text{m}^2/\text{N}$
Benzene	0.689	1.24	4550	95
Diethyl ether	0.922	1.66	8950	187
Ethyl alcohol	0.622	1.12	5310	111
Glycerin	0.281	0.505	1010	21
Heptane (n)	0.683	1.23	6890	144
Mercury	0.101	0.182	192	4.02
Water	0.115	0.207	2200	45.9
Source: Adapted by permission of the publisher from Zemansky, M. W., Abbott, M. M., Van Ness, H. C., 1975. Basic Engineering Thermodynamics, second ed. McGraw Hill, New York.

$β = 0.207 × 10^{−6} \text{K}^{−1}$

$κ = 45.9 × 10^{−11} \text{m}^2/\text{N}$

and, from Table C.1b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that

$v = v_f(20.0°\text{C}) = 0.001002 \text{m}^3\text{/kg}$

Then, from Eq. (11.30), we have

$c_p − c_v = \frac{Tβ^2v}{κ} = \frac{(293 \text{K})(0.207 × 10^{−6} \text{K}^{−1})^2 (0.001002 \text{m}^3\text{/kg})}{45.9 × 10^{−11} \text{m.s}^2\text{/kg}}$

$= 2.74 × 10^{−5} \text{J/(kg.K)} = 2.74 × 10^{−8} \text{KJ/(kg.K)}$

In most applications, this difference is clearly negligible, since the value of $c_p$ for liquid water at standard temperature and pressure is $4.18 \text{KJ/(kg.K)}$ .