Older freezers developed a coating of ice inside that had to be melted periodically; an electric heater could speed this defrosting process. Suppose you're melting ice from your freezer using a heating wire that carries a current of 3.4 when connected to 120 .What is the resistance of the wire?R=How long will it take the heater to melt 780 of accumulated ice at -10 ? Assume that all of the heat goes into warming and melting the ice, and that the melt water runs out and doesn't warm further.t=

Question

contestada

Respuesta :

klefenz klefenz · Answer 1 · 2019-09-09T01:04:02+02:00

Answer:

R = 35.3 ohms

It would take 11.2 minutes

Explanation:

Ohm's law states that

[tex]R = \frac{V}{I}[/tex]

Therefore:

[tex]R = \frac{120 V}{3.4 A} = 35.3 ohms[/tex]

Joule's law states that

[tex]P = I^2 * R[/tex]

So

[tex]P = 3.4^2 * 35.3 = 408 w[/tex]

I assume the problem means 780 g of ice.

The specific heat capacity of ice is:

Cp = 2 kJ/(kg*K)

And the latent heat to melt ice is:

Cl = 333 kJ/kg

So, the heat needed to melt ice from -10 C is:

Q = m * Cp * (tfinal - ti) + m * Cl

Q = 0.78 * (2 * (0 - (-10)) + 333) = 275 kJ

Power is energy over time

[tex]P = \frac{Q}{t}[/tex]

so with a power of P it would take t seconds to deliver an energy Q

[tex]t = \frac{Q}{P}[/tex]

[tex]t = \frac{275000}{408} = 674 s = 11.2 minutes[/tex]