Question

Estimate the rate at which ice would melt in a wooden box 20 mm thick having outside measurements (1 × 0.6 x 0.6) m assuming the external temperature to be 303 K. Thermal conductivity of wood = 0.168 W m-'K-!.​

Answer 1

Answer:

Area of all the six faces of box will conduct heat from out side to inside the box.

Here A=[4×200×100+2×100×100]

=10×104=105cm2

Δx=20mm=2cm,

K=0.0004cals−1cm−1K−1

Latent heat of ice L,=80cal/g

dQdt=mLΔT=KA(T1−T2)Δx

or mΔt=KA(T1−T2)ΔxL=0.0004×105×272×80=6.75g/s

Explanation:

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Answer 2

Explanation:

Area of all the six faces of box will conduct heat from out side to inside the box.

Here `A=[4xx200xx100+2xx100xx100]`

`=10xx10^(4)=10^(5) cm^(2)`

`Delta x=20 mm=2 cm`,

`K=0.0004 cal s^(-1) cm^(-1) K^(-1)`

Latent heat of ice `L,=80 cal//g`

`(dQ)/(dt)=(mL)/(Delta T)=(KA(T_(1)-T_(2)))/(Delta x)`

or `(m)/(Delta t)=(KA(T_(1)-T_(2)))/(Delta xL)=(0.0004xx10^(5)xx27)/(2xx80)=6.75 g//s`

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