Chapter 14, Problem 42P

Approximately how long should it take 8.2 kg of ice at 0°C to melt when it is placed in a carefully sealed Styrofoam ice chest of dimensions 25 cm x 35 cm x 55 cm whose walls are 1.5 cm thick? Assume that the conductivity of Styrofoam is double that of air and that the outside temperature is 34°C.

To determine

Time for ice to melt placed in a sealed Styrofoam ice chest.

Answer to Problem 42P

Solution:

$31.17\times {10}^5\text{s}$

Explanation of Solution

Given:

Mass of ice = 8.2 kg

Temperature of ice $=0^{\text{o}}\text{C = 273 K}$

Dimensions of ice chest = $\text{25 cm}\times \text{35 cm }\times \text{55 cm}$

Thickness of wall = 1.5 cm

Conductivity of Styrofoam is double that of air $\text{=}\text{2}\times \text{0}\text{.023 J/s}\text{m K = 0}\text{.046}\text{J/s}\text{m}\text{K}$.

Temperature of the outside = ${\text{34}}^{\text{o}}\text{C}\text{=}30\text{7}\text{K}$

Formula Used:

Rate of heat flow,

$\frac{\Delta Q}{\Delta t}=\frac{kA\left(T_2-T_1\right)}{l}$

Where,

k is the conductivity

T is the temperature

A is the area

l is the thickness

T is the time

Q is the heat

$Q=m{L}_f$

Where, m is the mass

L_f is the latent heat of ice.

Calculation:

Area of Styrofoam ice chest

$\begin{array}{l}2\left(l\times b+b\times h+l\times h\right)\\ =2\left(25\times 35+35\times 55+25\times 55\right){\text{cm}}^2\\ =8350{\text{cm}}^2=0.84{\text{m}}^2\end{array}$

$\begin{array}{l}Q=m{L}_f\\ =8.2\text{kg×3}{\text{.33×10}}^{\text{5}}\text{J/kg}\\ \text{=27}\text{.3}\times {\text{10}}^5\text{J}\end{array}$

$\begin{array}{l}\frac{\Delta Q}{\Delta t}=\frac{kA\left(T_2-T_1\right)}{l}\\ \Delta t=\frac{\Delta Q\times l}{kA\left(T_2-T_1\right)}=\frac{27.3\times {10}^5\text{J}\times 1.5\times {10}^{-2}\text{m}}{0.046\text{J/s}\text{m}\text{K}\times \text{0}\text{.84}{\text{m}}^2\times \left(307-273\right)\text{K}}\\ =31.17\times {10}^5\text{s}\end{array}$