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Nookie1986 [14]

2 years ago

7

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to 35 C from 175 C, its net radian

t power decreases to 12.0 W. What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at 22 C as the dish cools down.

1 answer:

pogonyaev2 years ago

3 0

Answer:

net radiant power is 275.28 W

Explanation:

Given data

net radiant power P(F) = 12 W

initial temperature Ta = 175 °C = 175 + 273 = 448 K

final temperature Tb = 35 °C = 35 + 273 = 308 K

temperature in the kitchen T(f) = 22 °C = 22 + 273 = 295 K

to find out

net radiant power

solution

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A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

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Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

3 0

2 years ago

What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface

PolarNik [594]

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

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Thus the gravitational force = 150 N

The correct option is a. 150 N

5 0

2 years ago

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

f = fo $\frac{v}{v-v)s}$ when the source moves towards the observer

f ’=f_o $\frac{v}{v+v_{sy}}$ Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

f’= f₀ $\frac{343}{343 + v_s}$

8 0

2 years ago

A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o

mariarad [96]

Answer:

$S_i=\frac{9}{10} =0.9$

Explanation:

Given:

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the volume level of oil when the ice melted, $V'=87\ cm^3$

<u>Now, therefore the volume of ice:</u>

$V_i=V-V_o$

$V_i=90-60$

$V_i=30\ cm^3$

<u>Now the volume of water:</u>

$V_w=V'-V_o$

$V_w=87-60$

$V_w=27\ cm^3$

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

$S_i=\frac{\rho_i}{\rho_w}$ .....................(1)

as we know that density is given as:

$\rm \rho=\frac{mass}{volume}$

now eq. (1)

$S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}$

where, m = mass of the water or the ice which remains constant in any phase

$S_i=\frac{V_w}{V_i}$

$S_i=\frac{27}{30}$

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7 0

2 years ago

A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is

astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B =(1.40 T)i

B =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)

F= 6.68*10¹¹⁴ N (-k)

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N

8 0

2 years ago