JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 7)
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 $$\times$$ 105 J kg$$-$$1, specific heat of ice = 2 $$\times$$ 103 J kg$$-$$1 and density of ice = 103 kg/m3
0.353 s
35.3 s
3.53 s
70.6 s
Explanation
Given, the length of the water pipe, L = 1 m
The cross-sectional area of the water pipe, A = 1 cm2 = 10$$-$$4 m2
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, Lf = 3.33 $$\times$$ 105 J/kg
The density of the ice, d = 1000 kg/m3
The specific heat of the ice, cp, ice = 2 $$\times$$ 103 J/kg
Heat required to melt the ice at 10$$^\circ$$C to 0$$^\circ$$C
Q = mcp$$\Delta$$T + mLf $$\Rightarrow$$ Q = dVcp$$\Delta$$T + dVLf
= 1000 $$\times$$ 10$$-$$4 $$\times$$ 2 $$\times$$ 103 $$\times$$ (10) + 1000 $$\times$$ 10$$-$$4 $$\times$$ 3.33 $$\times$$ 105 ($$\because$$ V = A $$\times$$ L)
= 35300 J
According to the Joule's law of heating,
H = I2Rt
$$\Rightarrow$$ 35300 = (0.5)2(4000) (t)
$$ \Rightarrow $$ t = 35.3 s
Thus, the minimum time required to melt the ice is 35.3 s.
The cross-sectional area of the water pipe, A = 1 cm2 = 10$$-$$4 m2
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, Lf = 3.33 $$\times$$ 105 J/kg
The density of the ice, d = 1000 kg/m3
The specific heat of the ice, cp, ice = 2 $$\times$$ 103 J/kg
Heat required to melt the ice at 10$$^\circ$$C to 0$$^\circ$$C
Q = mcp$$\Delta$$T + mLf $$\Rightarrow$$ Q = dVcp$$\Delta$$T + dVLf
= 1000 $$\times$$ 10$$-$$4 $$\times$$ 2 $$\times$$ 103 $$\times$$ (10) + 1000 $$\times$$ 10$$-$$4 $$\times$$ 3.33 $$\times$$ 105 ($$\because$$ V = A $$\times$$ L)
= 35300 J
According to the Joule's law of heating,
H = I2Rt
$$\Rightarrow$$ 35300 = (0.5)2(4000) (t)
$$ \Rightarrow $$ t = 35.3 s
Thus, the minimum time required to melt the ice is 35.3 s.
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