Due to cold weather a 1 m water pipe of cross-sectional area 1 cm^2 is filled with ice at -10°C. Res

Q: Why do we need to calculate the heat for warming the ice first?

Melting (phase change) only occurs at the melting point, which is $0^\circ\text{C}$ for water. Since the ice is initially at $-10^\circ\text{C}$, you must first supply sensible heat ($mc\Delta T$) to raise its temperature to $0^\circ\text{C}$ before you can supply latent heat ($mL$) to melt it.

Q: Why is the specific heat of ice different from the specific heat of water?

Specific heat depends on the molecular structure and the degrees of freedom of the substance. In solid ice, water molecules are locked in a rigid lattice, restricting their movement compared to liquid water, which means it takes a different amount of energy to raise its temperature.

Q: What if the resistance was given in ohms instead of kilo-ohms?

You must always convert all given values to standard SI units before calculating. Here, $4 \text{ k}\Omega$ must be converted to $4000 \Omega$ to ensure the calculated heat is in standard Joules.

Q: Does the shape of the pipe matter for this calculation?

No, the specific shape of the pipe does not matter. Only the total volume of the ice matters, which we found using the cross-sectional area and the length.

Q: What does 'assuming all heat produced is used for melting' mean?

It means we are assuming $100\%$ efficiency. In a real-world scenario, some of the heat generated by the resistor would be lost to the surrounding air or the pipe material itself, meaning it would take longer to melt the ice.

Question

Due to cold weather a $1 	ext{ m}$ water pipe of cross-sectional area $1 	ext{ cm}^2$ is filled with ice at $-10^\circ	ext{C}$. Resistive heating is used to melt the ice. Current of $0.5 	ext{ A}$ is passed through $4 	ext{ k}\Omega$ 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 $= 3.33 	imes 10^5 	ext{ J kg}^{-1}$, specific heat of ice $= 2 	imes 10^3 	ext{ J kg}^{-1}$ and density of ice $= 10^3 	ext{ kg/m}^3$]

Options:

(A) $0.353 	ext{ s}$

(B) $35.3 	ext{ s}$

(C) $3.53 	ext{ s}$

(D) $70.6 	ext{ s}$

Sigma Percentile · Accepted Answer

Imagine a water pipe frozen solid in the dead of winter. We have a $1 	ext{ m}$ long pipe filled with ice at $-10^\circ	ext{C}$. To melt it, we are passing an electrical current through a resistor placed inside the pipe. This problem beautifully combines the principles of calorimetry with electrical heating.

### Finding the Mass of the Ice

First, we need to know exactly how much ice we are dealing with. To find the mass of the ice, we simply multiply its density by its volume. The volume of a cylindrical pipe is the product of its cross-sectional area and its length.

$$ V = A 	imes L $$

Let's plug in the numbers. The density of ice is given as $1000 	ext{ kg/m}^3$. The area is $1 	ext{ cm}^2$, which we must convert to $10^{-4} 	ext{ m}^2$, and the length is exactly $1 	ext{ m}$.

$$ m = d 	imes V = 10^3 	imes (10^{-4} 	imes 1) $$

Multiplying these together, we find that there is exactly $0.1 	ext{ kg}$, or $100 	ext{ g}$, of ice inside the pipe.

### Calculating the Total Heat Required

Now, how much heat is needed to completely melt this ice? The heat goes into two distinct stages: first, warming the solid ice from $-10^\circ	ext{C}$ to $0^\circ	ext{C}$, and second, actually melting it into water at $0^\circ	ext{C}$.

$$ Q = mc_p \Delta T + mL_f $$

We substitute our calculated mass of $0.1 	ext{ kg}$, the specific heat of ice ($2 	imes 10^3 	ext{ J kg}^{-1} ^\circ	ext{C}^{-1}$), the temperature change of $10^\circ	ext{C}$, and the latent heat of fusion ($3.33 	imes 10^5 	ext{ J kg}^{-1}$) into our equation.

$$ Q = 0.1 	imes (2 	imes 10^3) 	imes 10 + 0.1 	imes (3.33 	imes 10^5) $$

The heat required to warm the ice is $2000 	ext{ J}$, and the heat to melt it is $33300 	ext{ J}$. Adding them up, we need a total of $35300 	ext{ J}$ of heat energy.

### Equating with Joule Heating

Where is this heat coming from? It's generated by the electrical resistance. According to **Joule's Law of Heating**, the heat produced is equal to the square of the current, multiplied by the resistance, and the time.

$$ H = I^2Rt $$

Since all the electrical heat is used to melt the ice, we equate our total heat required to the Joule heating formula. We plug in the current of $0.5 	ext{ A}$ and the resistance of $4000 \Omega$.

$$ 35300 = (0.5)^2 	imes 4000 	imes t $$

$0.5$ squared is $0.25$. Multiplying that by $4000$ gives us $1000$. Dividing $35300$ by $1000$, we get our final answer:

$$ t = 35.3 	ext{ s} $$

In reality, some heat would escape to the surroundings, so it would take slightly longer. But assuming perfect efficiency, it takes just over half a minute!

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Visualized Solution

$Visualizing the Setup$

$Mass of the Ice$

$Substituting Values for Mass$

$Calculating Mass$

$Total Heat Required$

$Substituting Values for Heat$

$Calculating Total Heat$

$Joule Heating$

$Equating Heat Energies$

$Solving for Time$

$The Way Forward$

The Sigma Insight: Calorimetry

Finding the Mass of the Ice

Calculating the Total Heat Required

Equating with Joule Heating

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Ice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Heat Required

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Heat

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Heat

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Joule Heating

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Heat Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Time

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward