Question

A hollow spherical iron container whose outer diameter is \(40 \mathrm{cm}\) and thickness is \(0.4 \mathrm{cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(3^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, and the rate at which ice melts in the container.

Short answer

Answer: The approximate rate of heat loss from the iron container is around 7539.8 W. The ice melts at a rate of approximately 0.0225 kg/s inside the container.

Step-by-step solution

Key Parameters and Constants

Before starting to solve the problem, we should list all the key parameters and constants we'll need to use. They are: Outer diameter of the container, \(D = 40\) cm Thickness of the container, \(t = 0.4\) cm Outer surface temperature, \(T_{out} = 3^{\circ} \mathrm{C}\) Temperature of the iced water, \(T_{in} = 0^{\circ} \mathrm{C}\) Furthermore, let's note the constants: Specific heat capacity of iron, \(c_p = 449 \ J/kg °C\) Thermal conductivity of iron, \(k = 80 \ W/m °C \) Latent heat of fusion of ice, \(L_f = 334 \times 10^3 \ J/kg\) Also, note that the container is hollow, so the inner radius and outer radius are different.

Finding the Inner and Outer Radii

Since the outer diameter is given, we can calculate the outer radius. Outer radius, \(R_{out} = \frac{D}{2} = 20\) cm Next, we need to find the inner radius. Inner radius, \(R_{in} = R_{out} - t = 20 - 0.4 = 19.6 \) cm In order to use the thermal conductivity formula, we need the radii in meters. Outer radius, \(R_{out}(m) = \frac{20}{100} = 0.2\ m\) Inner radius, \(R_{in}(m) = \frac{19.6}{100} = 0.196\ m\)

Calculate the Rate of Heat Loss

We'll use the formula for heat conduction through a sphere to calculate the rate of heat loss: \(q = 4 \pi (k) \frac{R_{out} R_{in}}{R_{out} - R_{in}} \Delta T\) where \(q\) is the rate of heat flow or heat loss, \(k\) is the thermal conductivity of iron, \(R_{out}\) and \(R_{in}\) are the outer and inner radii, and \(\Delta T\) is the temperature difference. Plugging in the values: \(q = 4 \pi (80) \frac{(0.2)(0.196)}{0.2 - 0.196} (3 - 0)\) \(q = 320 \pi \cdot 0.0025 \cdot (3-0) = 2400 \pi \ W\) Therefore, the rate of heat loss is approximately \(7539.8\ W\).

Calculate the Rate at Which Ice Melts

To find the rate at which ice melts in the container, we'll use the formula: \(Q = mL_f\) where \(Q\) is the heat absorbed by the ice, \(m\) is the mass of the ice, and \(L_f\) is the latent heat of fusion of ice. We will assume the rate of heat loss from the sphere equals the heat absorbed by the ice. Therefore, \(Q = 2400 \pi\ W\). Now, we can find the rate of ice melting by dividing the heat absorbed by the latent heat of fusion: Rate of ice melting \(= \frac{Q}{L_f} = \frac{2400 \pi}{334 \times 10^3} \frac{W}{J/kg}\) This gives us: Rate of ice melting \(\approx 0.0225\ \frac{kg}{s}\) So, the ice is melting at a rate of approximately \(0.0225\ \frac{kg}{s}\).