A Styrofoam cooler has a surface area of $2,700 \mathrm{~cm}^{2}$ and a wall thickness of $3.0 \mathrm{~cm}$. Styrofoam has a thermal conductivity of $0.010 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$. A $2.0 \mathrm{~kg}$ block of ice is placed inside the cooler that has a temperature inside of $2.0^{\circ} \mathrm{C}$. If the heat of fusion for ice is $333.7 \mathrm{~kJ} / \mathrm{kg}$ and the temperature outside is $35.0^{\circ} \mathrm{C}$, then how long will the ice last? Ignore the effect of the air inside the cooler.

Question

Quizplus · Accepted Answer

The rate of heat transfer ($Q/t$) through the walls of the cooler can be calculated using the formula $Q/t = kA(T_{\text{outside}} - T_{\text{inside}})/d$, where $k$ is the thermal conductivity of Styrofoam, $A$ is the surface area, $T_{\text{outside}} - T_{\text{inside}}$ is the temperature difference between the outside and inside of the cooler, and $d$ is the thickness of the cooler walls. Substituting the given values, $Q/t = 0.010 \times 2700 \times (35 - 2)/0.03$, converting the area to $m^2$ and the thickness to meters. The energy required to melt the ice is given by $Q = m \times L_f$, where $m$ is the mass of the ice and $L_f$ is the heat of fusion. Substituting the given values, $Q = 2.0 \times 333,700$. Dividing the total energy required to melt the ice by the rate of heat transfer gives the time in seconds, which is then converted to hours.

A Styrofoam Cooler Has a Surface Area Of 2,700 cm22,700 \mathrm{~cm}^{2}2,700 cm2 And a Wall Thickness Of

A Styrofoam Cooler Has a Surface Area Of $2,700 \mathrm{~cm}^{2}$ And a Wall Thickness Of