Question

Define coefficient of thermal conductivity of a material. A cubical thermocole box of side 20 cm
and wall thickness 4 cm is full of ice. If outside temperature is 40 °C, estimate the amount of
ice melted in five hours (K for thermocol is 0.01 J s–1 °C–1 and latent heat of fusion of ice is 335 J
g–1)

Answer 1

Answer on Question #53696, Physics Mechanics Kinematics Dynamics

Define coefficient of thermal conductivity of a material. A cubical thermocole box of side $20~\mathrm{cm}$ and wall thickness $4~\mathrm{cm}$ is full of ice. If outside temperature is $40{}^{\circ}\mathrm{C}$, estimate the amount of ice melted in five hours (K for thermocol is $0.01~\mathrm{J}~\mathrm{s}^{-1}$ ${}^\circ \mathrm{C}-1$ and latent heat of fusion of ice is $335~\mathrm{J}~\mathrm{g}^{-1}$)

Solution

The quantity of heat flowing into the ice through all six faces of the box is given by Eq.(1)

$$Q = \frac {K A \left(\theta_ {1} - \theta_ {2}\right) t}{x}$$

If $m$ be the mass of the ice melted due to this heat and $L$, the heat of fusion of water, then $Q = mL$.

Thus,

$$m L = \frac {K A \left(\theta_ {1} - \theta_ {2}\right)}{x}$$

Here, the area of each face of cubical box $(side)^2 = 20 \times 20 = 400cm^2 = 0.04m^2$

The total area of all six faces of the cubical box, $A = 6 \cdot 0.04 = 0.24m^2$

Thickness of each face, $x = 0.04m$

The time for which heat flows, $t = 5hr = 5\cdot 60\cdot 60 = 18000\mathrm{sec}$

The heat of fusion of water $L = 335J / gm$

Thermal conductivity of thermocole, $K = 0.01J / (\sec \cdot m \cdot {}^{0}C)$

The temperature difference, $\theta_{1} - \theta_{2} = 40 - 0 = 40^{0}C$

Mass of the ice melted,

$$m = \frac {K A \left(\theta_ {1} - \theta_ {2}\right) t}{x L} = \frac {0 . 0 1 \cdot 0 . 2 4 \cdot 4 0 \cdot 1 8 0 0 0}{0 . 0 5 \cdot 3 3 5} = 1 0 3. 1 6 g m = 0. 1 0 3 k g$$

Answer: $m = \frac{KA(\theta_1 - \theta_2)t}{xL} = 0.103kg$

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