Solution for Question 26
Analysis:
In the steady state, there is a continuous heat flow from the hot end (16°C) to the cold end (-40°C). The water near the cold piston freezes into ice. An interface exists at $0^\circ\text{C}$.
1. Heat Flux Continuity:
The rate of heat flow ($H$) through the ice must equal the rate of heat flow through the water.
$$H_{ice} = H_{water}$$
Using the formula $H = \frac{KA\Delta T}{L}$:
$$\frac{K_{ice} A [0 – (-40)]}{L_{ice}} = \frac{K_{water} A [16 – 0]}{L_{water}}$$
Given $K_{ice} = 4 K_{water}$, we substitute:
$$\frac{4 K_{water} (40)}{L_{ice}} = \frac{K_{water} (16)}{L_{water}}$$
$$\frac{160}{L_{ice}} = \frac{16}{L_{water}} \Rightarrow \frac{10}{L_{ice}} = \frac{1}{L_{water}} \Rightarrow L_{ice} = 10 L_{water}$$
2. Mass Conservation:
The total mass of the substance remains constant. Initially, the pipe (length 80 cm) is full of water.
$$M_{total} = \rho_{water} A (80 \text{ cm})$$
In the final state:
$$M_{total} = M_{ice} + M_{water} = \rho_{ice} A L_{ice} + \rho_{water} A L_{water}$$
Equating the masses (and canceling $A$):
$$900 L_{ice} + 1000 L_{water} = 1000(80)$$
Substitute $L_{ice} = 10 L_{water}$:
$$900(10 L_{water}) + 1000 L_{water} = 80000$$
$$9000 L_{water} + 1000 L_{water} = 80000$$
$$10000 L_{water} = 80000 \Rightarrow L_{water} = 8 \text{ cm}$$
Then $L_{ice} = 10 \times 8 = 80 \text{ cm}$.
Total distance = $L_{ice} + L_{water} = 80 + 8 = 88 \text{ cm}$.