Question

I place a 500-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm2 .

If the air surrounding the box is at 20°C and after 4 hours the ice is completely melted, what is the conductivity of the Styrofoam material? (Lf = 80 cal/g)
a. 9.6 × 10−5cal/s⋅cm⋅°C
b. 2.8 × 10−6cal/s⋅cm⋅°C
c. 1.15 × 10−2cal/s⋅cm⋅°C
d. 2.3 × 10−4cal/s⋅cm⋅°C

Answer 1

Para resolver este problema es necesario aplicar los conceptos relacionados a la conductividad térmica, para lo cual se tiene matematicamente que

`k=(\Delta Q)/(\Delta t) (1)/(A) (x)/(\Delta T)`

Where,

K = thermal conductivity

A = Cross-sectional Area

`\Delta T`= Change at temperature

x = Distance

`\Delta t`= Difference of time

`\Delta Q` = Heat exchange energy

Our values are given as follow,

`t=4hours ((3600s)/(1hour)) = 14400s`

The total heat required to change the phase of ice would be

`H = L_f*m \rightarrow` Where Lf Latent heat of fussion and m is the mass.

`H = 80*500g`

`H =40000 cal`

`A= 600cm^2`

`\Delta T = 20\° C`

`x = 1cm`

Replacing we would have:

`k=(\Delta Q)/(\Delta t) (1)/(A) (x)/(\Delta T)`

`k=(40000)/(14400) (1)/(600) (1)/(20)`

`K = 2.3*10^(-4) cal/s\cdot cm\cdot \°C`

Therefore the correct answer is D.