A 0.316-m-thick sheet of ice covers a lake. The airtemperature at the ice surfac
ID: 1724245 • Letter: A
Question
A 0.316-m-thick sheet of ice covers a lake. The airtemperature at the ice surface is -17.5 °C. In 3.10 minutes,the ice thickens by a small amount. Assume that no heat flows fromthe ground below into the water and that the added ice is very thincompared to 0.316 m. Calculate the number of millimeters by whichthe ice thickens. I don't understand how to solve this. Please show me step bystep how to solve this. A 0.316-m-thick sheet of ice covers a lake. The airtemperature at the ice surface is -17.5 °C. In 3.10 minutes,the ice thickens by a small amount. Assume that no heat flows fromthe ground below into the water and that the added ice is very thincompared to 0.316 m. Calculate the number of millimeters by whichthe ice thickens. I don't understand how to solve this. Please show me step bystep how to solve this.Explanation / Answer
This is a conduction problem. . Mass of new ice is density ofice times thickness of new ice times area of new ice: . m = s A where s is what we want to solve for. . The rate of heat flow through the ice sheet is given bythe conduction equation: . H = k A T / d where d = 0.316 meters . Also, k is the conductivity of ice (in atable in your book) and T is 17.5 deg C . The rate of heat flow is heat per time, andthe heat is the amount of heat lost by the water that then froze,i.e.: . Q = mLf where Lf is the latent heat of fusion forwater. . Put all this together and we get: . Q / t = k A T /d m Lf / t = k A T /d . s A Lf / t = k A T /d eliminate A on both sides and solve for s: . s = t kT / d Lf = . = 186 * 2 * 17.5 / 0.316 * 917 * 333000 = 6.7465x 10-5 meters = 0.0675mm note: it makes sense that this is a verysmall number, since the time is only 3 minutes. This is about 1.3 mm per hour, which would add about inch per day tothe ice sheet.Related Questions
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