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Thermal Properties of Matter

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Question : 19 of 22
Marks: +1, -0
A ‘thermocole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermocole is 0.01Js−1m−1K−10.01 \mathrm{J} \mathrm{s}^{-1} \mathrm{m}^{-1} \mathrm{K}^{-1}.
[Heat of fusion of water =335×103Jkg−1]= 335 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1}]
Solution:  
Here, length of each side, l=30cm=0.3ml = 30 \mathrm{cm} = 0.3 \mathrm{m}
Thickness of each side, Δx=5cm=0.05m\Delta x = 5 \mathrm{cm} = 0.05 \mathrm{m}
total surface area through which heat enters into the box,
A=6l2A=6 l^{2}
=6×0.3×0.3=0.54m2=6 \times 0.3 \times 0.3=0.54 \mathrm{m}^{2}
Temp. diff.,ΔT=45−0=45∘C\Delta T = 45 - 0 = 45^{\circ}\mathrm{C},
K=0.01Js−1m−1K−1K=0.01 \mathrm{J} \mathrm{s}^{-1} \mathrm{m}^{-1} \mathrm{K}^{-1}
time, Δt=6hrs=6×60×60s\Delta t=6 \mathrm{hrs}=6 \times 60 \times 60 \mathrm{s}
Latent heat of fusion, L=335×103Jkg−1L=335\times 10^{3} \mathrm{J} \mathrm{kg}^{-1}
Let mm be the mass of ice melted in this time
∴ΔQ=mL=KA(ΔTΔx)Δt\therefore \Delta Q = m L = K A \left( \frac{\Delta T}{\Delta x} \right) \Delta t
m=KA(ΔTΔx)ΔtLm=K A \left( \frac{\Delta T}{\Delta x} \right) \frac{\Delta t}{L}
=0.01×0.54×450.05×6×60×60335×103=0.01 \times 0.54 \times \frac{45}{0.05} \times \frac{6 \times 60 \times 60}{335 \times 10^{3}}
=0.313kg=0.313 \mathrm{kg}
Mass of ice left =4−.313=3.687kg= 4 - .313 = 3.687 \mathrm{kg}
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