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kaheart [24]
3 years ago
7

Substance A has twice the specific heat capacity as substance B. If 1000 J of heat are added to 1.0 kg of each substance, compar

e the change in temperature of each substance.
Physics
1 answer:
Mrrafil [7]3 years ago
4 0
Substance A would have a delta T (change in temp) rise 1/2 the rise in substance B.

Q=mc x delta T

Q= heat energy in Joules
m= mass of substance heated or cooled
c= specific heat
ΔT is change in temp.

Solve for change in temp=. Q/mc

Specific heat and mass are not inversely proportional to heat energy input.

Putting into real world scenario of using water to heat a building.

Specific heat of water is 1.
It takes 1 btu to raise one pound of water 1 degF. at a base of 60 degF

Acetone specific heat is .51

So it takes half the amount of heat input to get a 100 degree ΔT, as compared to water.
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Archy [21]
The balloon will shrink because the average kinetic energy of gas molecules in a balloon decreases with fall in temperature. Butif we warm the balloon , it will rise.

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3 0
2 years ago
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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
Which of the following is a scalar quantity?
neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0
2 years ago
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A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag
olga2289 [7]
Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.
8 0
3 years ago
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DedPeter [7]
I think it might be A. I’m sorry if I’m wrong
5 0
3 years ago
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