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Step-by-Step Solution
Step 1: List the Given Data
1. Length of the pipe, $L = 1\text{ m}$
2. Cross-sectional area, $A = 1\text{ cm}^2 = 10^{-4}\text{ m}^2$
3. Initial temperature of ice, $T_i = -10^\circ\text{C}$
4. Final temperature (melting point of ice), $T_f = 0^\circ\text{C}$
5. Current, $I = 0.5\text{ A}$
6. Resistance, $R = 4\,000\text{ }\Omega = 4\text{ k}\Omega$
7. Latent heat of fusion of ice, $L_f = 3.33 \times 10^5\text{ J kg}^{-1}$
8. Specific heat capacity of ice, $c = 2 \times 10^3\text{ J kg}^{-1} \text{K}^{-1}$
9. Density of ice, $\rho = 1000\text{ kg m}^{-3}$
Step 2: Calculate the Volume and Mass of Ice
The volume of ice in the pipe is given by
$V = A \times L = (10^{-4}\,\text{m}^2) \times (1\,\text{m}) = 10^{-4}\,\text{m}^3.$
Hence, the mass of ice is
$m = \rho \times V = 1000\,\text{kg m}^{-3} \times 10^{-4}\,\text{m}^3 = 0.1\,\text{kg}.$
Step 3: Determine the Heat Required to Raise Ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$ and Melt It
The total heat $Q$ needed has two parts:
1. Heat to warm the ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$.
2. Heat to melt the ice completely at $0^\circ\text{C}$.
Mathematically,
$$
Q = m\,c\,\Delta T + m\,L_f,
$$
where $\Delta T = (0 - (-10)) = 10^\circ\text{C}.$
Substituting values:
$$
Q = 0.1 \times (2 \times 10^3) \times 10 + 0.1 \times (3.33 \times 10^5).
$$
Calculate each term:
Heating from $-10^\circ\text{C}$ to $0^\circ\text{C}$:
$0.1 \times 2 \times 10^3 \times 10 = 2000\text{ J}.$
Melting ice at $0^\circ\text{C}$:
$0.1 \times 3.33 \times 10^5 = 3.33 \times 10^4\text{ J}.$
Hence,
$$
Q = 2000 + 3.33 \times 10^4 = 3.53 \times 10^4\text{ J}
$$
(which is $35{,}300\text{ J}$).
Step 4: Use Joule’s Law of Heating to Find the Time
The electrical energy supplied by the resistor is given by Joule’s law:
$$
H = I^2 R \, t.
$$
We want all this energy to equal $Q,$ which is the heat required to melt and warm the ice.
Therefore,
$$
3.53 \times 10^4 = (0.5)^2 \times 4000 \times t.
$$
Simplify:
$$
(0.5)^2 = 0.25,\quad \quad 0.25 \times 4000 = 1000.
$$
Hence,
$$
3.53 \times 10^4 = 1000 \, t.
$$
Solving for $t$:
$$
t = \dfrac{3.53 \times 10^4}{1000} = 35.3 \text{ s}.
$$
Step 5: Final Answer
Therefore, the minimum time required to melt the ice completely using the given resistive heating setup is
$35.3\text{ s}.$
