What is meant by the term resistance?

Physics

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

Definiton 1.the refusal to accept or comply with something; the attempt to prevent something by action or argument

Definition 2. the ability not to be affected by something, especially adversely

Definition 3. the impeding, slowing, or stopping effect exerted by one material thing on another

Definiton 4: the degree to which a substance or device opposes the passage of an electric current, causing energy dissipation. Ohm's law resistance (measured in ohms) is equal to the voltage divided by the current.

Explanation:

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ( $Q$ ), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ( $U_{l,ice}$ , $U_{l,steam}$ ), in joules, and two sensible component ( $U_{s,w}$ ), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

$c_{ice}$ - Specific heat of ice, in joules per kilogram per degree Celsius.

$c_{w}$ - Specific heat of water, in joules per kilogram per degree Celsius.

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$ , $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$ , $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$ , $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$ , then the amount of thermal energy is: