The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TB
b) TA < TB
c) TA = TB
d) More information is needed
Answers
The final temperatures are such that TA > TB.
The specific heat capacity refers to the quantity of heat required to raise the temperature of 1 Kg of a body by 1K. The higher the specific heat capacity of a body, the higher the quantity of heat required to raise the temperature of the body and vice versa.
Hence, if the specific heat of substance A is greater than that of substance B and A and B are at the same initial temperature, when equal amounts of energy are added to them, the final temperature are such that TA > TB.
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Answer:
For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:
d) More information is needed
(The relation between the masses is not given)
Explanation:
For this case we know the following info:
Where c means specific heat for the substance A and B.
We also know that the initial temperatures for both sustances are equal:
We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:
And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:
And if we replace the formula for sensible heat we got:
And if we replace for the change of the temperature we got:
And since we have this:
For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:
d) More information is needed
(The relation between the masses is not given)
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Answers
Answer:
The vertical trajectory is governed by Ordinary Differential Equation.
Time derivatives of each state variables.
d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.
Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.
Therefore,
F = -mg - D + T, If V is positive and
F = -mg + D - T, If T is negative.
D is drag and the questions gave it as zero.
Explanation:
The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p
Answers
The molar latent enthalpy of boiling of iron at 3330 K is ΔH = 342 10^3 J.
Explanation:
Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.
d ln p = (ΔH / RT^2) dt
(1/p) dp = (ΔH / RT^2) dt
dp / dt = p (ΔH / RT^2) = 3.72 10^-3
(p) (ΔH) / (8.31) (3330)^2 = 3.72 10^-3
ΔH = 342 10^3 J.
Answers
The work required to bring the sphere to stop is equal to the kinetic energy possessed by the sphere.
Kinetic energy of a rotating body is given by,
K.E =
Here, I= Moment of inertia of hollow sphere,
Since, the hollow sphere is rotating about the axis passing through its center, I =
M= Mass of the sphere= 3.8 kg,
R= Radius of gyration= Radius of the sphere= 0.25 m
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Kinetic energy =
= 13.4 J
∴ Work required to bring the sphere to stop is 13.4 J.
Answers
ANSWER:
D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures
STEP-BY-STEP EXPLANATION:
One of the main reasons fusion power cannot be harnessed is that its power requirements are incredibly high. For fusion to occur, a temperature of at least 100,000,000°C is needed.
Therefore, the correct answer is D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures
Answers
Answer:
Lever, pulley and wheel and axle are the types of machine it's grouped to
Explanation:
The wheel and axle is a simple machine that works by reducing friction in trying to move a load. This is seen in the Tyre of the bicycle
Pulley is a simple machine that creates a mechanical advantage and supports the changing of direction for a rope or cable. This is seen in the chain of the bicycle
Levers attached to the bike's pedals are pushed down to direct force into the pulley system.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answers
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225