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Anna71 [15]
3 years ago
12

Which of the following best describes a property of water?

Physics
2 answers:
Snezhnost [94]3 years ago
5 0

Answer: C. Remaining liquid over a wide temperature range

Explanation: Water has a high specific heat i.e it requires a large amount of heat to raise the temperature of 1 gram of water through 1^0C. This is because the molecules of water are associated through strong hydrogen bonds.

It is also an universal solvent as it can dissolve most of the substances in it due to its high dielectric constant.

The freezing point of water is  0^0C, that is liquid water converts to solid form at 0^0C and the boiling point of water is  100^0C, that is liquid water converts to gaseous form at 100^0C. Thus it can remain in liquid form over the temperature range of  0^0C to 100^0C.

Water has a high surface tension as the molecules of water are held together by strong hydrogen bonds.

stiks02 [169]3 years ago
3 0

Correct answer is option C.

Remaining liquid over a wide temperature range  

Other options are invalid.

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A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium. True or
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Read 2 more answers
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
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