Problem 86 - Master Resource Book in JEE Main Physics

Question

One end of a copper rod of length \(1.0 \mathrm{~m}\) and area of cross-section \(10^{-3} \mathrm{~m}^{2}\) is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is \(92 \mathrm{cal} / \mathrm{m}-\mathrm{s}-{ }^{\circ} \mathrm{C}\) and the latent heat of ice is \(8 \times 10^{4} \mathrm{cal} / \mathrm{kg}\), then the amount of ice which will melt in one min is (a) \(9.2 \times 10^{-3} \mathrm{~kg}\) (b) \(8 \times 10^{-3} \mathrm{~kg}\) (c) \(6.9 \times 10^{-3} \mathrm{~kg}\) (d) \(5.4 \times 10^{-1} \mathrm{~kg}\)

Step-by-Step Solution

Answer

The amount of ice that will melt in one minute is \(6.9 \times 10^{-3} \mathrm{~kg}\).

Step 1: Determine Temperature Difference

The copper rod transfers heat from the boiling water to the ice. The temperature of boiling water is approximately 100°C, and the temperature of ice is 0°C. Thus, the temperature difference, \( \Delta T \), is 100°C - 0°C = 100°C.

Step 2: Use Fourier's Law of Heat Conduction

Fourier's law for heat conduction states \( \frac{dQ}{dt} = -k \cdot A \cdot \frac{dT}{dx} \), where \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, and \( \frac{dT}{dx} \) is the temperature gradient. Substituting the known values, the heat flow \( \frac{dQ}{dt} = -92 \times 10^{-3} \times \frac{100}{1} = 9.2 \mathrm{~cal/s} \). Here, \( k = 92 \mathrm{~cal/m-s-°C} \), \( A = 10^{-3} \mathrm{~m}^{2} \), \( \Delta T = 100 \mathrm{~°C} \), and \( L = 1 \mathrm{~m} \).

Step 3: Calculate Total Heat Conducted in One Minute

Since the heat is conducted continuously over one minute (or 60 seconds), multiply the heat flow by time: \( Q = \frac{dQ}{dt} \times 60 = 9.2 \times 60 \mathrm{~cal} = 552 \mathrm{~cal} \).

Step 4: Determine Amount of Ice Melted

To find the amount of ice melted, use the formula \( Q = m \cdot L \), where \( m \) is the mass of the ice that melts and \( L \) is the latent heat of fusion for ice. Rearranging gives \( m = \frac{Q}{L} = \frac{552}{8 \times 10^4} \).

Step 5: Simplify the Calculation

Calculate the fraction: \( m = \frac{552}{8 \times 10^4} = 6.9 \times 10^{-3} \mathrm{~kg} \). Substitute \( Q = 552 \mathrm{~cal} \) and \( L = 8 \times 10^4 \mathrm{~cal/kg} \).

Key Concepts

Fourier's Law of Heat Conduction

Imagine you're trying to understand how heat travels through a solid object. Fourier's Law of Heat Conduction is a key principle that explains this process. It states that the rate of heat transfer through a material is directly proportional to the negative gradient of temperatures and area through which the heat flows. This can be written as:

• \( \frac{dQ}{dt} = -k \cdot A \cdot \frac{dT}{dx} \)

Here, \( dQ/dt \) is the rate of heat transfer (in cal/s), \( k \) is the thermal conductivity of the material (copper, in this case), \( A \) is the cross-sectional area through which heat flows, and \( dT/dx \) is the temperature gradient. This law helps us calculate how much heat moves across the rod over time. In our exercise, substituting the known values of \( k \), \( A \), and \( \Delta T \) gives us the rate of heat flow. This is crucial for understanding how much heat is moving through the copper rod. The negative sign indicates that heat flows from a hotter region to a cooler one, which is why the heat moves from boiling water to ice.

Temperature Gradient

The temperature gradient is the change in temperature over a specific distance. It’s like a temperature map along the rod. In simpler terms, it shows how temperature varies from one end of the rod to the other. Imagine one end of the copper rod in boiling water (100°C) and the other end in ice (0°C). The difference is 100°C over 1 meter. This gives a temperature gradient of \( \frac{100 °C}{1 \text{ m}} \), or simply 100°C/m.

• This gradient is critical because it dictates the speed and direction of heat transfer.

A larger gradient means faster heat movement. It sets up the condition for applying Fourier's Law, informing us how efficiently the rod conducts heat from one end to the other.

Latent Heat of Fusion

When heat is transferred to the ice, it doesn’t just increase the temperature right away. Instead, it works to change the state from solid (ice) to liquid (water). The energy required to do this is called the latent heat of fusion. For ice, this value is \( 8 \times 10^4 \text{ cal/kg} \).

• Latent heat is all about the energy needed to transition between phases without changing temperature.

In our scenario, the heat transferred (calculated with Fourier's law) is used to melt the ice. To determine how much ice melts, we use the formula \( Q = m \cdot L \), where \( L \) represents the latent heat of fusion. This brings home the idea that to melt a specific amount of ice, it requires energy not just to warm it, but to actually force a change in its phase.

Heat Transfer in Solids

Understanding heat transfer in solids involves knowing how heat energy moves from one part of the solid to another. In solids like our copper rod, this process is governed by thermal conductivity, the material’s ability to conduct heat. Solids conduct heat better due to closely packed atoms that easily pass kinetic energy to one another.

• This type of heat transfer doesn’t rely on the movement of material, solely on the transfer of kinetic energy between particles.

In the copper rod example, heat travels from the hot end (immersed in boiling water) to the cold end (surrounding the ice) - this direct heat transfer is a typical case of conduction through solids. The efficiency in heating transfer is determined by the thermal conductivity of copper (92 cal/m-s-°C). Recognizing these properties of solid materials is essential to predicting and controlling how much heat will pass through, affecting how much ice melts in a given time.