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ANTONII [103]
2 years ago
8

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

hool and in first-year college physics courses, the formula is usually given asQ ΔH m Q mCp ΔT mCpT2 ␣T1 (2)T1(a) What assumption about Cp is required to go from Equation 1 to Equation 2? (b) The heat capacity (Cp) of liquid n-hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well-insulated vessel containing 2.00L of liquid n–C6H14 at T 300 K. A combustion reaction known to release 16.73 kJ of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be 3.10 K. In a separate experiment, it is found that 6.14kJ of heat is required to raise the temperature of everything in the system except the hexane by 3.10 K. Use these data to estimate Cp[kJ/(mol K)] for liquid n-hexane at T 300 K, assuming that the condition required for the validity ofEquation 2 is satisfied. Compare your result with a tabulated value.
Engineering
1 answer:
a_sh-v [17]2 years ago
6 0

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

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The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,
sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity = \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

T_i = 25 degree C = 298 k

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diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226

\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}

\tau = 0.036 t

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Ai = 0.982

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\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}

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7 0
2 years ago
A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal
Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)

10*2=\sqrt{1+8F^{2} } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_{1} }

Here, g = gravity = 32.2 ft/s²

v=7.4162*\sqrt{32.2*1} =42.0833ft/s

b) The flow rate:

q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s

c) The Froude number:

v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s

F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345

d) The flow energy dissipated:

E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft

e) The critical depth:

d_{c} =(\frac{(\frac{q}{L})^{2}  }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2}  }{32.2} )^{1/3} =3.8030ft

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