Thick ness of ice on a lake is 5 cm. and the temperature of air is `-20^(@)C.` If the rate of cooling of water inside the lake be `20000 cal mi n^(-1)` through each square metre surface, find K for ice.

Answer

`[3.5 Wm^(-1) .^(@)C^(-1)]`

Explanation

Here, `Delta x=5 cm=5xx10^(-2)m`,
`Delta T=0-(-20)=20^(@)C`
`(Delta Q)/(Delta t)=20000"cal "mi n^(-1)=(20000)/(60)cal. s^(-1)`
`=(20000)/(60)xx4.2 j s^(-1) =1400 j s^(-1)`
`A=1 sq m, K=?`
As, `(Delta Q)/(Delta t)=KA((Delta T)/(Delta x))` ltbbrgt `:. K=(Delta Q//Delta t)/(A(Delta T//Delta x))=(1400)/(1xx20//(5xx10^(-2))) =3.5 Wm^(-1) .^(@)C^(-1)`