For the circuit shown in the figure (Figure 1) a 20.0 Ω resistor is embedded in a large block of ice at 0.00∘C, and the battery has negligible internal resistance.

At what rate (in g/s) is this circuit melting the ice? (The latent heat of fusion for ice is 3.34×105J/kg.)

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aristocles aristocles · Answer 1 · 2020-04-10T19:57:00+02:00

Answer:

rate of melting of ice is given as

[tex]\frac{dm}{dt} = 3.56 \times 10^{-2} g/s[/tex]

Explanation:

As we know that 20 ohm , 10 ohm and 20 ohm resistor is in parallel with each other

so the equivalent of the all three is given as

[tex]\frac{1}{R} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10}[/tex]

so we have

[tex]R = 5 ohm[/tex]

now total resistance of whole circuit is given as

[tex]R = 15 + 5 + 5 + 10[/tex]

now current flowing in the circuit is given as

[tex]i = \frac{45}{35} = \frac{9}{7} A[/tex]

now current in 20 ohm resistor is given as

[tex]i_1 = \frac{30}{50} (\frac{9}{7})[/tex]

[tex]i_1 = 0.77 A[/tex]

now we know that power dissipated across the resistor is used to melt the ice

so we have

[tex]\frac{dm}{dt} L = i_1^2 R[/tex]

[tex]\frac{dm}{dt} (3.34 \times 10^5) = (0.77)^2 (20)[/tex]

[tex]\frac{dm}{dt} = 3.56 \times 10^{-5} kg/s[/tex]

so in required units it is

[tex]\frac{dm}{dt} = 3.56 \times 10^{-2} g/s[/tex]