I place a 600-g ice cube (initially at 0°C)in a Styrofoam box with wall thickness 1.8 cm and total surface area 700 cm².If the air surrounding the box is at 25°C and after 6 hours the ice is completely melted,what is the conductivity of the Styrofoam material? (L_f = 80 cal/g)

Question

Quizplus · Accepted Answer

We can use the formula Q = kATΔT/Δx to solve for the conductivity k, where Q is the heat transfer rate (which we can set equal to the energy required to melt the ice), A is the total surface area of the box, T is the time the ice takes to melt (6 hours in this case), ΔT is the temperature difference between the ice and the surroundings (25°C - 0°C = 25°C), and Δx is the thickness of the Styrofoam wall.

First, we need to find the energy required to melt the ice. This is simply:

Q = (600 g) x (80 cal/g) = 48,000 cal

Next, we can plug in the given values and solve for k:

48,000 cal = k(700 cm^2)(6 hours)(25°C)/(1.8 cm)

k = 2.4 x 10^-14 cal/s·cm·°C

Converting to scientific notation with correct units, we get D) 2.4 x 10^-14 cal/s·cm·°C.

I Place a 600-G Ice Cube (Initially at 0°C)in a Styrofoam