A lake is covered with ince 2 cm thick.t he temperature of ambient air is `-15^(@)C` find the rate of thickening of ice. For ice `k=4xx10^(-4)k-cal-m^(-1)s^(_1)(.^(@)C)^(-1)`. Density `=0.9xx10^(3)kg//m^(3)` and latent heat `L=80` kilo cal/kg

A lake is covered with ice 2 cm thick.the temperature of ambient air is -15^(@)C find the rate of thickening of ice. For ice k=4xx10^(-4)k-cal-m^(-1)s^(_1)(.^(@)C)^(-1) . Density =0.9xx10^(3)kg//m^(3) and latent heat L=80 kilo cal/kg

Calculate the rate of increment of the thickness of ice layer on a lake when thickness of ice is 10 cm and the air temperature is -5^(@) C . If thermal conductivity of ice is 0.008 cal "cm"^(-1) "s"^(-1)°"C"^(-1) , density of ice is 0.91 xx 10^(3) "kg m"^(-3) and latent heat is 79.8 cal "gm"^(-1) . How long will it take the layer to become 10.1 cm ?

The thickness of ice in a lake is 5cm and the atmospheric temperature is 10^(@)C Calculate the time rquired for the thickness of ice to grow to 7cm Therml conductivity for ice =4 xx 10^(-3) cal cm^(-1) s^(-1) c^(-10 density of ice =0.92 g //cm^(3) and latent heat of fusion of ice =80 cal//g .

A lake is covered with ice 5 cm thick and the atmospheric temperature above the ice is – 10^(@)C . At what rate (in cm/hour) will the ice layer thicken? Thermal conductivity of ice = 0.005 cgs unit, density of ice = 0.9 g//cc and latent heat of fusion of ice = 80 cal/g.

The thickness of ice on a Jake is 6 cm and the temperature of air is - 10^(@) C . At what rate is the thickness of ice increasing ? Thermal conductivity of ice = 2 "Wm"^(-1) "K"^(-1) . Density of ice = 920 "kg m"^(3) , specific latent heat of ice = 336 "KJ kg"^(-1)

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)

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)