Solve the Fiendish Physics Final Exam Challenge

[sasuke07]

Homework Statement

A fiendish physics instructor devises the following exercise for students as a pass/fail final exam for their heat course. A 110 g cube of ice (5 cm on a side) is placed in a form fitting Styrofoam box that is 2 mm thick. The ice cube and the inside of the box are at a temperature of 0 degrees C. The entire box is now sealed and submerged in a bath of boiling water (100 degrees C) which maintains a constant temperature on the outside of the box. IF the student opens the box before the ice has totally melted he fails the course. IF the ice has melted but the temperature of the water is more than 2 degrees C he fails. Some of the following numbers may be helpful: coefficient of thermal expansion of ice =51E-6 (1/C); thermal conductivity of ice = 2.2 J/s-m-C; thermal conductivity of Styrofoam = 0.01 J/s-m-C; latent heat of fusion of water = 33.5E4 J/kg; heat capacity of water =4.18 kJ/kg-C;

Homework Equations

change in Q=(KAChange in T Change in t)/L
Change in Q=cmChange in T,

The Attempt at a Solution

I have a few questions, i was wondering if the coefficeint of thermal expansion of the ice was the same as coeffiecient of thermal conductivity, is the termoconductivity of styrofaom important? and is the latent heat of fusion of water important?, I just basically don't really know where to start
After plugging in the known numbers i got C
Change in Q=4.18*11kg*Change in T
Change in Q=(K1.5m*100C*Cahnge in t)/.002m
I am assuming that heat capactity of water and heat capacity of ice are the same, and that the surface area is 6*5^2 because a cube has 6 sides)

 

[Andrew Mason]

The Attempt at a Solution

I have a few questions, i was wondering if the coefficeint of thermal expansion of the ice was the same as coeffiecient of thermal conductivity, is the termoconductivity of styrofaom important? and is the latent heat of fusion of water important?, I just basically don't really know where to start
After plugging in the known numbers i got C
Change in Q=4.18*11kg*Change in T
Change in Q=(K1.5m*100C*Cahnge in t)/.002m
I am assuming that heat capactity of water and heat capacity of ice are the same, and that the surface area is 6*5^2 because a cube has 6 sides)

Divide the problem into two parts:

1. heat flow required to melt the ice.

2. heat flow required to raise the temperature of the water by 2°C.

You have to work out the rate at which heat flows into the box. What determines that?

Knowing how long it takes to melt the ice and how long after that it takes to raise the temperature 2°C will give you a time "window" in which to open the box.

AM