All categories

1 month ago

What is the surprising thing that happens in a superconductor?

1 answer:

8 0

Generally, the electrical resistance of nearly all materials declines when they are subjected to lower temperatures. As they cool down further, this resistance continues to decrease. However, for some unique materials known as 'superconductors', an extraordinary phenomenon occurs: when their temperature decreases to a certain threshold, their resistance vanishes entirely, reaching absolute zero rather than just a negligible figure. In practical terms, if you initiate a current in a superconducting wire and complete the circuit by connecting the wire ends, that current can circulate indefinitely, maintaining its flow without diminishing for extended periods, provided the wire remains sufficiently chilled. This absence of resistance is key for various applications. For instance, numerous CT and MRI scanning devices depend on robust electromagnets, which consist of coils made from superconducting wires, kept in liquid helium. While this setup may sound intricate and costly, it's actually more practical and economical than using conventional electromagnets with copper wire and the substantial power supplies required for their operation. Additionally, wire resistance leads to energy loss when current moves through it. Energy dissipates along the lengthy cables from power plants to homes, resulting in unusable energy that electric companies cannot sell. Concepts are being introduced to create superconducting cables that transport electricity efficiently from power plants to consumers. These cables would be designed as hollow tubes insulated to maintain low temperatures with liquid helium or hydrogen inside, thus eliminating energy loss during transmission. Even though this approach could involve complexity and cost, the anticipated benefits, in terms of energy efficiency, outweigh the losses currently incurred with standard cables.

You might be interested in

Which of the following could be considered a scientific statement?

Anarel [2989]

Answer: D) X-rays exhibit lower frequency than microwaves

Explanation:

A scientific statement is one that relies on evidence gathered from experiments and direct observation of the natural world. Such a statement can be corroborated through appropriate experimental methods.

The option D illustrates a scientific statement because it allows for the frequency of both microwaves and X-rays to be assessed and contrasted. This process entails experimental and observational techniques, which validates it as a scientific statement.

6 0

2 months ago

Read 2 more answers

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

1 month ago

In this lab, you will do experiments to identify types of changes. Using the question format you learned (shown above), write an

VMariaS [2998]

Response:

How can you differentiate a physical change from a chemical change?

Clarification:

4 0

2 months ago

Which statement explains why sulfur is classified as a Group 16 element?

Tems11 [2777]

Answer: The correct option is (1).

Explanation:

Group 16 is the third-from-last column in the periodic table and is known as the oxygen family.

Members of this group include oxygen, sulfur, selenium, tellurium, and polonium.

Elements in Group 16 have 6 valence electrons in their outermost shell.

The electronic configuration of sulfur is $1s^22s^22p^63s^23p^4$ .

Because sulfur belongs to Group 16, it has 6 valence electrons.

3 0

1 month ago

Read 2 more answers

BRAINLIESTTT ASAP!!!

lorasvet [2795]

Although I may not be the smartest, I can definitely answer.

This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.

8 0

3 months ago

Read 2 more answers