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andriy [413]
3 years ago
12

In this experiment, you need to examine the idea of thermal energy transfer. Using a controlled experiment, what might a good qu

estion about the variables that affect thermal energy transfer be? Thermal energy transfer depends on many properties, but limit your question to only two.
Physics
3 answers:
Rama09 [41]3 years ago
7 0

Answer:

We shall consider two properties:

1. Temperature difference

2. Thermal conductivity of the material

Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.

One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.

This experiment will show that

(a) heat flows from a region of high temperature to a region of lower temperature.

(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.

(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.

Explanation:

USPshnik [31]3 years ago
5 0

Answer:

the effect would be 2w=btyh,3

Explanation:

Jayy3 years ago
0 0

answer: How do mass and the type of material affect thermal energy transfer
on edg its right

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you have been provided with an insulated wire , voltmeter or a battery with a charge of 10ohms, a bulb, a switch and a resistor
eduard
I’m trying to get my points up I have the same question
4 0
3 years ago
Se tienen 500g de alcohol etílico a una temperatura de -40 °C ¿Cuánto calor se necesita para transformarlo a vapor a una tempera
Feliz [49]

The question is: You have 500g of ethyl alcohol at a temperature of -40 ° C. How much heat is needed to transform it into steam at a temperature of 150ºC?

Answer: 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of -40^{o}C to 150^{o}C.

Explanation:

Given: Mass = 500 g

Initial temperature = -40^{o}C

Final temperature = 150^{o}C

The standard value of specific heat of ethyl alcohol is 2.46 J/g ^{o}C.

Formula used to calculate the heat energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass of substance

C = specific heat

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = m \times C \times (T_{2} - T_{1})\\= 500 g \times 2.46 J/g^{o}C \times [150 - (-40)]^{o}C\\= 233700 J

Thus, we can conclude that 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of -40^{o}C to 150^{o}C.

8 0
3 years ago
What are the developmental stages according to Erikson?
natita [175]

Answer:

1. Trust vs. Mistrust

2. Autonomy vs. Shame

3. Initiative vs. Guilt

4. Industry vs. Inferiority

8 0
3 years ago
Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0
Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}

j-direction:

m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J

Answer: The system lost 500J worth of kinetic energy in the collision

4 0
3 years ago
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0
3 years ago
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