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

Respuesta :

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

[tex]k=\frac{\Delta Q}{\Delta t} \frac{1}{A} \frac{x}{\Delta T}[/tex]

Where,

K = thermal conductivity

A = Cross-sectional Area

[tex]\Delta T[/tex]= Change at temperature

x = Distance

[tex]\Delta t[/tex]= Difference of time

[tex]\Delta Q[/tex] = Heat exchange energy

Our values are given as follow,

[tex]t=4hours (\frac{3600s}{1hour}) = 14400s[/tex]

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

[tex]H = L_f*m \rightarrow[/tex] Where Lf Latent heat of fussion and m is the mass.

[tex]H = 80*500g[/tex]

[tex]H =40000 cal[/tex]

[tex]A= 600cm^2[/tex]

[tex]\Delta T = 20\° C[/tex]

[tex]x = 1cm[/tex]

Replacing we would have:

[tex]k=\frac{\Delta Q}{\Delta t} \frac{1}{A} \frac{x}{\Delta T}[/tex]

[tex]k=\frac{40000}{14400} \frac{1}{600} \frac{1}{20}[/tex]

[tex]K = 2.3*10^{-4} cal/s\cdot cm\cdot \°C[/tex]

Therefore the correct answer is D.

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