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allsm [11]
3 years ago
11

Remove the ice cube tray and the plastic bag from the freezer and let them sit for 3 minutes. Then gently remove the piece of ic

e from the tray and the piece of ice from the plastic bag. Place each piece of ice on a piece of foil. Check the pieces of ice about every three minutes until you can clearly see whether there’s a difference in how fast they are melting. What do you observe? HELPPP HURRYYYY
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

That when you leave the ice setting without foil, it melts slower. When you put the piece of ice in the foil it melts faster than just letting it set without foil. I really hope this helps.

Explanation:

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
6. State whether each of the following is an example of conduction, convection, and/or radiation. Explain if you think there is
hodyreva [135]
A).  Convection is heating the soup in the pot.
When you stick the spoon into the hot soup,
conduction heats the spoon all the way up to the end.

b).  Water conducts heat a little bit. 
But convection is much more responsible for the
uniform distribution of temperature in the kiddie pool.

c).  The heat from the metal bench conducts directly
to the buttus epidermis when you sit on it.

d).  You feel the heat on your face ... but not on the back of your
neck ... on account of radiation from the fire and the hot grill.
3 0
3 years ago
What types of materials can make up sediments?
GREYUIT [131]

Answer: Pieces of minerals, rocks, plant and animal remains.

Explanation: Pieces of minerals, rocks, plant and animal remains. whether or not patterns cause flow rates of rivers to vary. sand settles from faster-moving water;smaller costs of silt and clay that form up mud settle from slower moving water.

4 0
3 years ago
A pump, submerged at the bottom of a well that is 35 m deep, is used to pump water uphill to a house that is 50 m above the top
Sidana [21]

Answer:

0.45

Explanation:

6 0
3 years ago
During each cycle, a refrigerator ejects 610 kJ of energy to a high-temperature reservoir, and takes in 505 kJ of energy from a
jenyasd209 [6]

Answer

A. the work done on the refrigerant in each cycle is 105kJ

B the coefficient of performance of the refrigerator is 4.8

Explanation

Given data

Work done at high temperature T2 Qh=610kJ

Work done at low temperature T1 Ql=505kJ

We know that the net work done by the refrigerator is expressed as

Wnet= Qh-Ql

=610-505

=105kJ

Also we know that the coefficient of performance is expressed as

COP= Ql/Wnet

COP= 505/105

= 4.8

8 0
3 years ago
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