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=

R = 35.3 ohms

It would take 11.2 minutes

Explanation:

Ohm's law states that

Therefore:

Joule's law states that

So

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

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

car a and b are moving in opposite directions

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