A physical thermal problem, solved in college physics thermal problem

Let the resistors of A and B be r1 and r2 respectively, and the heat required for water heating to boiling is q A: 220 2 r1*t1=q (1).

A and B in series: 220 2 (r1+r2)*t2=q (2) A and B in parallel: 220 2*(r1+r2) (r1*r2)*t3=q (3).

1) (2), get r1*t1=(r1+r2)*t2 launch r1:r2=t2:(t1-t2) (4)1) (3), get r1*t1=(r1*r2) (r1+r2)*t3, so t3=t1(r1+r2) r2

Substitute (4) into the above equation to get.

t3=t1*(t2+t1-t2)/(t1-t2)=t1^2/(t1-t2)

The first question, give points.

Water is the same, so the work used to heat the water is also the same, and the voltage is also the same, so the formula w=u 2 r*t is used to calculate, according to the title.

w=u 2 r1*t1=u 2 (r1+r2)*t2, so r2=(t2 t1-1)r1

When connected in parallel, r=r1r2 (r1+r2) is brought into w=u 2 r*t3.

t3=（t2-t1）t1/t2

The three pots of water are the same, so the work required is the same p1=w t1=u 2 r1, p2=w t2=u 2 r2, and p3=u 2 ((r1r2) (r1+r2))=u 2 r1+u 2 r2=p1+p2, so t3=w (p1+p2)=t1t2 (t1+t2).

Summary. 1.If the heat capacity c of a system is a constant parameter, then its heat q can be expressed by the following formula as its temperature t rises:

q = c * t2.If the heat capacity c of a system is a variable parameter, then its heat q can be expressed by the following formula when its temperature t rises: q = c(t) *t

1.If the heat capacity c of a system is a constant parameter, then its heat q can be expressed by the following formula when its temperature t rises: orange q = c * t2

If the heat capacity c of a system is a variable parameter, then its thermal state Q can be expressed by the following formula: q = c(t) *t when its temperature t rises

Container A holds 1mol of ideal gas at an initial state temperature of 300K. It is connected by a thin tube (and a closed valve) to Vessel B, which is 4 times the volume of Vessel A, which in its initial state holds the same amount of ideal gas as Xiang Changhui at 300 K degrees of Wenshen Xunshan. The valve opens so that the pressure equalizes and the temperature of each container is the same in the final state.

The in-swim entropy increases in the final state of all 2mol gases.

Seek entropy increase. When the valve is opened, the entropy increase of 2 mol of ideal gas in container A and container B is the same as that of the canopy. Because they equalize the pressure, the temperature will also equalize, so their final state temperature is also the same.

Since the temperature is the same, the entropy increase of 2mol gas is also the same.

Find the numeric value. Both Container A and Container B have a final temperature of 300K and the same pressure.

h=ms=

Therefore, the initial velocity of the wooden block when it leaves the table is 5m s (it should be needless to say that the flat throwing motion should be said?). All kinetic energy is 25j

The kinetic energy converted into internal energy is.

So the conversion rate is available.

The first question is the process of bullet being fired into a wooden block.

Momentum is conserved, mv0=(m+m)v

Then do a flat toss together.

t=root(2hg)=

v=s/t=1m/s

v0=(m+m)v/m=201m/s

The heat generated by the bullet into the wooden block q = mv0 2 2 - (m+m)*v 2 2 = 201j

The amount of heat absorbed by the bullet q'=cm△t=

Absorption rate = q'q = 78 percent

Explanation of the second problem: Assuming that the molecular flow and the wall of the organ act for the time t, the mass of the molecule that acts with the wall during this time.

m=vstn0

m0 is obtained from the momentum theorem.

ft=2mv

So f=2n0

m0sv2 is obtained from the pressure formula.

p=f/s=2n0

m0v2

In physics, "the sum of the kinetic energy of all the molecules in the object moving irregularly and the energy of the molecules is called the internal energy of the object", please put forward a reasonable conjecture on the "factors affecting the internal energy of the object" based on the knowledge you have learned, give a reasonable conclusion, and explain the basis for reasoning (refer to the example).Example:

The magnitude of the internal energy of an object may be related to the temperature of the object. Inferential conclusion: The higher the temperature of an object, the greater the internal energy. Reasoning basis: The higher the temperature, the faster the molecular movement and the greater the molecular kinetic energy. 1）

The magnitude of the internal energy of the object may also be related to the mass of the object. 2) Inferential conclusion: the greater the mass of the object, the more energy there is in the object

3) Reasoning basis: The larger the mass, the greater the number of molecules, the more the sum of the kinetic energy and molecular energy of all the molecules in the object to move irregularly4) Talk about your understanding of internal energy and mechanical energy.

Internal energy is the energy determined by the microscopic particles of the object, and mechanical energy is the energy determined by the macroscopic two parts of the object as a whole.

The internal energy of an object is related to the mass of the object.

Conclusion: The greater the mass, the greater the internal energy.

Basis: The greater the mass, the more molecules, the greater the molecular potential energy and kinetic energy.

Mechanical energy is from a macroscopic object, it is related to the kinetic energy and potential energy of the object, the kinetic energy is determined by the mass and velocity, and the potential energy is related to the mass, position, deformation.

Internal energy is microscopic and is determined by the number of molecules and their kinetic and potential energy.

1) The magnitude of the internal energy of the object may also be related to the volume of the object. 2) Inferential conclusion: [The volume of the object increases, and the potential energy of the molecule also increases].

3) Reasoning basis: [The volume of the object is larger, the distance between the molecules increases, and the molecular potential energy increases].4) Talk about your understanding of internal energy and mechanical energy.

Object internal energy is a microscopic representation of the energy inside an object. Mechanical energy is the energy on the macroscopic surface of an object

At room temperature and pressure, saturated water vapor, moist air and dry air can be well approximated as ideal gases, i.e., pv=nrt=m m *rt, p=m v *rt m= m *rt, =pm (rt), m= rt p

1.It can be calculated from the saturated water vapor density o=:saturated vapor pressure p1,p1= *2.

The partial pressure of water vapor in moist air = , the partial pressure of air = 1*10 The partial pressure is proportional to the mass n, assuming that the total mass of the matter is 1mol, where the number of moles of water vapor is n1 and the air is 1-n1Easy to obtain: n1=, 1-n1=

3.Mean molar mass of moist air =

4.Density of moist air = pm (rt), density ratio to dry air =

Moist air = dry air + water vapor.

For water vapor:

The mass of moist air (relative humidity of 90%) is 90% of that of saturated water vapor and the volume is the same, so the density:

1=ρ0*90%= kg/m^3

For dry air:

Air quality in 1mol: m= kg

p0v0=nrt

10^5*v0=1*

v0= m^3

Air density: 2 = m v0 = kg m 3

Density ratio of moist air (relative humidity of 90%) and dry air = (1 + 2) 2 =

The density ratio of moist air (relative humidity of 90%) and dry air is: 1