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` .

Latent heat of fusion of ice : 80 cal/g :: Steam :…………..

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)

Complete the analogy: Latent heat of fusion of ice : 80 cal/g :: steam:

A lake surface is exposed to an atmosphere where the temperature is lt 0^(@)C . If the thickness of the ice layer formed on the surface grows form 2cm to 4cm in 1 hour. The atmospheric temperature, T_(a) will be- (Thermal conductivity of ice K = 4 xx 10^(-3) cal//cm//s//.^(@)C , density of ice = 0.9 gm//ice. Latent heat of fustion of ice = 80 cal//m . Neglect the change of density during the state change. Assume that the water below the ice has 0^(@) temperature every where)

A lake surface is exposed to an atmosphere where the temperature is lt 0^(@)C . If the thickness of the ice layer formed on the surface grows form 2cm to 4cm in 1 hour. The atmospheric temperature, T_(a) will be- (Thermal conductivity of ice K = 4 xx 10^(-3) cal//cm//s//.^(@)C , density of ice = 0.9 gm//ice. Latent heat of fustion of ice = 80 cal//m . Neglect the change of density during the state change. Assume that the water below the ice has 0^(@) temperature every where)

The thickness of ice in a lake is 10 cm and its temperature is -10^(0)C . The thermal conductivity of ice = 0.004 cal cm^(-1)" sec"^(-1)""^(0)C^(-1) . Density of ice = 0.92 g/cc, latent heat of ice = 80 cal/g. The time taken for thickness of ice to be doubled is