Home
Class 11
PHYSICS
A ‘thermacole’ icebox is a cheap and eff...

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 co-efficient of thermal conductivity of thermacole is `0.01 J s^(–1) m^(–1) K^(–1)`. [Heat of fusion of water `= 335xx 103 J kg^(–1)]`

Text Solution

Verified by Experts

Area of six faces of the box,
`A=6l^(2)=6xx(0.3)^(2)=0.54m^(2)`
thickness, L = 5 cm = 0.05 m, time, t = 6 h = `6xx3600s`
temperature difference = `45-0=45^(@)C`
Total heat entering in to the box in 6 hours.
`Q=(KA(theta_(1)-theta_(2))t)/L=(0.01xx0.54xx45xx6xx3600)/0.05=104976J`
If .m. be the amount of ice melted, then
`m=Q/L_("ice")=(104976)/(335xx10^(3))=0.313kg`
`therefore` mass of ice left after `6h=4-0.313=3.687kg`.
Promotional Banner

Similar Questions

Explore conceptually related problems

A thermocole cubical icebox of side 30 cm has a thickness of 5.0 cm if 4.0 kg of ice are put ini the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45^@C and coefficient of thermal conductivity of thermocole =0.01 J//kg .

A cubical ice box of side 50 cm has a thickness of 5.0 cm. if 5 kg of ice is put in the box, estimate the amount of ice remaining after 4 hours. The outside temperature is 40^(@)C and coefficient of thermal conductivity of the material of the box = 0.01 Js^(-1) m^(-1) .^(@)C^(-1) . Heat of fusion of ice = 335 Jg^(-1) .

A cold box in the shape of a cube of edge 50 cm is made of ‘thermocol’ material 4.0 cm thick. If the outside temperature is 30^(@)C the quantity of ice that will melt each hour inside the ‘cold box’ is : (the thermal conductivity of the thermocol material is 0.050 W//mK)

A brass boiler has a base area of 0.15 "m"^(2) and thickness is 1.0 "cm" . It boils water at the rate of 6.0 "kg/min" . When placed on a gas stove. Estimate the temperature of the part of the flame in contact. With the boiler. Thermal conductivity of brass = 109 "Wm"^(-1) "K"^(-1)

A brass boiler has a base area of 0.15m^2 and thickness 1.0 cm it boils water at the rate of 6.0 kg//min , When placed on a gas. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass =109 J//s-.^@C ) and heat of vapourization of water =2256 J//g .

A brass boiler has a base area of 0.15m^2 and thickness 1.0 cm it boils water at the rate of 6.0 kg//min , When placed on a gas. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass =109 J//s-.^@C ) and heat of vapourization of water =2256 J//g .

One kg of ice at 0^(@)C is mixed with 1 kg of water at 10^(@)C . The resulting temperature will be

Time in which a layer of ice of thickness 10 cm will increase by 5 cm on te surface of a pond when temperature of the surrounding is -10^(@)C coefficient of thermal conductivity of ice K=0.005((Cal)/(cm-sec-.^(@)C)) and density rho=0.9gram//cm^(3) (L=80cal/g)

A slab of stone of area of 0.36 m^(2) and thickness 0.1 m is exposed on the lower surface to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = 3.63 xx 10^(5) J kg^(-1) )

Two vessels of different materials are identical in size and wall-thickness. They are filled with equal quantities of ice at 0^(@)C . If the ice melts completely in 10 and 25 minutes respectively, compare the coefficients of thermal conductivity of the materials of the vessels.