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Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and tot

Oksi-84 [34.3K]

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

5 0

3 years ago

Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have

Lera25 [3.4K]

Answer:

The extent of greenhouse effect on mars is $G_m = 87 K$

Explanation:

From the question we are told that

The albedo value of Mars is $A_1 = 0.15$

The albedo value of Mars is $A_2 = 0.15$

The surface temperature of Mars is $T_1 = 220 K$

The surface temperature of Venus is $T_2 = 700 K$

The distance of Mars from the sun is $d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m$

The distance of Venus from sun is $d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m$

The radius of the sun is $R = 7*10^{8} \ m$

The energy flux is $E = 6.28 * 10^{7} W/m^2$

The solar constant for Mars is mathematically represented as

$T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ]$

Where $\sigma$ is the Stefan's constant with a value $\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}$

So substituting values

$T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}$

$T = 307K$

So the greenhouse effect on Mars is

$G_m = T - T_1$

$G_m = 307 - 220$

$G_m = 87 K$

3 0

4 years ago

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p

asambeis [7]

Answer:

The volume of water that was in the kettle is 1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$ , then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$ $c_{s}$ (100 - 20 ) + $m_{w}$ $c_{w}$ (100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$

4 0

4 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

b)

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

4 years ago

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

o-na [289]

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

$F=ma$

$100=100m\times a$

$ma=1\ N$

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

$F_{100-99}=ma$

$F_{100-99}=1$

We need to calculate the total number of masses attached to the string

Using formula for mass

$m'=(100-50)m$

$m'=50m$

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

$F_{50}=m'a$

Put the value into the formula

$F_{50}=50m\times a$

$F_{50}=50\times1$

$F_{50}=50\ N$

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

8 0

4 years ago