A thin walled copper sphere of radius 5 cm and mass 100 g containing 100 g of water is cooled to - 176 °C by immersing it in liquid air. It is then placed inside a filling hollow sphere of expanded polythene of outer radius 10 cm in a room at 20 °C What is the value of thermal conductivity of ice if the ice just melts after 24 hours? (Specific heat capacities of ice and copper are 2.1 kJ/kgK and 0.4 kJ/kgK respectively and the specific latent heat of fusion of ice is 336 kJ/kg).

Answer 1

To find the thermal conductivity of ice using the information provided, we need to analyze the heat transfer process occurring when the cooled copper sphere (with the water inside) is placed in the polythene hollow sphere at room temperature.

Parameters Given:

1. Copper Sphere:

   • Radius \( r_c = 0.05 , m \)
   • Mass \( m_c = 0.1 , kg \)
   • Specific heat capacity of copper \( c_c = 0.4 , kJ/kgK = 400 , J/kgK \)
   • Initial temperature \( T_{c,i} = -176 , °C = 97 , K \)

2. Water Inside Copper Sphere:

   • Mass \( m_w = 0.1 , kg \)
   • Specific heat capacity of water \( c_w = 4.18 , kJ/kgK = 4180 , J/kgK \)
   • Initial temperature \( T_{w,i} = -176 , °C = 97 , K \)

3. Outer Sphere (Polythene):

   • Outer radius \( r_p = 0.1 , m \)
   • Inner radius (same as copper sphere) \( r_{p,i} = 0.05 , m \)
   • Room temperature \( T_{room} = 20 , °C = 293 , K \)

4. Latent heat of fusion of ice:

   • \( L_f = 336 , kJ/kg = 336000 , J/kg \)

5. Specific heat capacity of ice:

   • \( c_i = 2.1 , kJ/kgK = 2100 , J/kgK \)

6. Time:

   • \( t = 24 , hours = 86400 , seconds \)

Heat Transfer to Ice

When the copper sphere and water reach a temperature that's able to melt ice, the amount of heat lost by the copper and the water will equal the amount of heat gained by the ice.

Total heat lost by the copper and water can be calculated as follows:

1. Heat lost by copper when it warms to 0 °C:

   \[ Q_c = m_c \cdot c_c \cdot (0 - T_{c,i}) \] \[ Q_c = 0.1 \cdot 400 \cdot (0 - (-176)) = 0.1 \cdot 400 \cdot 176 = 7050 , J \]

2. Heat lost by water when it warms to 0 °C:

   \[ Q_w = m_w \cdot c_w \cdot (0 - T_{w,i}) \] \[ Q_w = 0.1 \cdot 4180 \cdot (0 - (-176)) = 0.1 \cdot 4180 \cdot 176 = 73680 , J \]

3. Total heat lost: \[ Q_{lost} = Q_c + Q_w = 7050 + 73680 = 80730 , J \]

4. Heat needed to melt ice: Assuming a mass \( m \) of ice melts after receiving the total heat:

   \[ Q_{gained} = m \cdot L_f \]

5. Setting heat lost equal to heat gained: \[ 80730 = m \cdot 336000 \] \[ m = \frac{80730}{336000} = 0.240 , kg \]

Thermal Conductivity Calculation

To find the thermal conductivity \( K \) of ice, we use the formula for heat transfer through a spherical shell:

\[ Q = \frac{K \cdot A \cdot (T_{room} - 0)}{(r_{p} - r_{p,i})} \cdot t \]

Where:

• \( A = 4\pi r_{p}^2 \) is the surface area of the outer sphere.
• \( r_p = 0.1 , m \) (outer radius of polythene sphere)
• \( r_{p,i} = 0.05 , m \) (inner radius of polythene sphere)
• \( T_{room} = 20 , °C = 293 , K \)
• Time \( t = 86400 , seconds \)

Calculating \( A \):

\[ A = 4\pi (0.1)^2 = 4\pi (0.01) = 0.12566 , m^2 \]

So now we express \( Q \):

\[ 80730 = \frac{K \cdot 0.12566 \cdot (293 - 273)}{0.1 - 0.05} \cdot 86400 \]

We need to find \( K \):

\[ 80730 = \frac{K \cdot 0.12566 \cdot 20}{0.05} \cdot 86400 \]

\[ 80730 = \frac{K \cdot 0.12566 \cdot 20 \cdot 86400}{0.05} \]

\[ 80730 = \frac{K \cdot 0.12566 \cdot 20 \cdot 86400}{0.05} \implies 80730 \cdot 0.05 = K \cdot 0.12566 \cdot 20 \cdot 86400 \]

Calculating the constant:

\[ 4036.5 = K \cdot 0.12566 \cdot 20 \cdot 86400 \]

Calculating \( 0.12566 \cdot 20 \cdot 86400 = 21696.96 \):

\[ 4036.5 = K \cdot 21696.96 \]

Now isolating \( K \):

\[ K = \frac{4036.5}{21696.96} \approx 0.186 , W/mK \]

Result

The value of thermal conductivity of ice is approximately:

\[ \boxed{0.186 , W/mK} \]