Question 219: Current Electricity

Question

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at $$-$$10$$^\circ$$C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 k$$\Omega$$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 $$	imes$$ 105 J kg$$-$$1, specific heat of ice = 2 $$	imes$$ 103 J kg$$-$$1 and density of ice = 103 kg/m3

Accepted Answer

Given, the length of the water pipe, L = 1 m
The cross-sectional area of the water pipe, A = 1 cm
2
= 10
$$-$$4
m
2
The temperature of the ice = $$-$$ 10$$^\circ$$C
Current passing in the conductor, I = 0.5 A
Resistance of the conductor, R = 4 k$$\Omega$$
The latent heat of fusion for ice, L
f
= 3.33 $$	imes$$ 10
5
J/kg
The density of the ice, d = 1000 kg/m
3
The specific heat of the ice, c
p, ice
= 2 $$	imes$$ 10
3
J/kg
Heat required to melt the ice at 10$$^\circ$$C to 0$$^\circ$$C
Q = mc
p
$$\Delta$$T + mL
f
$$\Rightarrow$$ Q = dVc
p
$$\Delta$$T + dVL
f
= 1000 $$	imes$$ 10
$$-$$4
$$	imes$$ 2 $$	imes$$ 10
3
$$	imes$$ (10) + 1000 $$	imes$$ 10
$$-$$4
$$	imes$$ 3.33 $$	imes$$ 10
5
($$\because$$ V = A $$	imes$$ L)
= 35300 J
According to the Joule's law of heating,
H = I
2
Rt
$$\Rightarrow$$ 35300 = (0.5)
2
(4000) (t)
$$ \Rightarrow $$ t = 35.3 s
Thus, the minimum time required to melt the ice is 35.3 s.

Answer

0.353 s

Answer

35.3 s

Answer

3.53 s

Answer

70.6 s

Current Electricity

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