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प्रश्न
A ‘thermacole’ 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 coefficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]
उत्तर १
Side of the given cubical ice box, s = 30 cm = 0.3 m
Thickness of the ice box, l = 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m = 4 kg
Time gap, t = 6 h = 6 × 60 × 60 s
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermacole, K = 0.01 J s–1 m–1 K–1
Heat of fusion of water, L = 335 × 103 J kg–1
Let m’ be the total amount of ice that melts in 6 h.
The amount of heat lost by the food: `theta = (KA(T-0)t)/l`
Where
A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3
`theta = (0.01xx0.54xx(45)xx6xx60xx60)/0.05 = 104976 J`
But `theta = m'L`
`:. m' = theta/L`
`= 104976/(335xx10^3) = 0.313 kg`
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.
उत्तर २
Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.
∴Total heat entering the icebox in 6 h is given by
`Q = (KA(T1 - T2)t)/d`
`= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"`
`= 1.05 xx10^5 J`
Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus
`1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)`
or `m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg`
The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is = (4.0 - 0.313) kg
= 3.687 kg
उत्तर ३
Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.
∴Total heat entering the icebox in 6 h is given by
`Q = (KA(T1 - T2)t)/d`
`= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"`
`= 1.05 xx10^5 J`
Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus
`1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)`
or `m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg`
The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is = (4.0 - 0.313) kg
= 3.687 kg
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