Ask question

Ask question

All categories

3 years ago

8

Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when

cooled from 90.7 to 39.4°C, as the heat provided when 182 g of steam is cooled from 118°C to 100.°C. (Assume that the specific heat of liquid water is 4.187 J/g·°C over the given temperature range. Also assume that the specific heat of water vapor above its boiling point is 2.078 J/g·°C.)

1 answer:

Nitella [24]3 years ago

3 0

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

You might be interested in

A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c

monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0

3 years ago

g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

4 years ago

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)

vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0

3 years ago

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

Paraphin [41]

Answer:

$10.8\ \text{lb/ft^2}$

$101.96\ \text{lb/ft}^2$

Explanation:

$v_1$ = Velocity of car = 65 mph = $65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}$

$\rho$ = Density of air = $0.00237\ \text{slug/ft}^3$

$v_2=0$

$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

$P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}$

The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

$P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2$

The maximum pressure on the girl's hand is $101.96\ \text{lb/ft}^2$

5 0

3 years ago

What is the 16 diget code to downlod pokemon ultra sun for free on ds

Kazeer [188]

Answer:

2468962218527611

Explanation:

7 0

3 years ago

Read 2 more answers