Question

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.)

Answer 1

Answer:

rate of melting of ice is given as

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

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

$\frac{1}{R} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10}$

so we have

$R = 5 ohm$

now total resistance of whole circuit is given as

$R = 15 + 5 + 5 + 10$

now current flowing in the circuit is given as

$i = \frac{45}{35} = \frac{9}{7} A$

now current in 20 ohm resistor is given as

$i_1 = \frac{30}{50} (\frac{9}{7})$

$i_1 = 0.77 A$

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

so we have

$\frac{dm}{dt} L = i_1^2 R$

$\frac{dm}{dt} (3.34 \times 10^5) = (0.77)^2 (20)$

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

so in required units it is

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