Ice cream removed from a freezer has a temperature of 35F Te

Ice cream removed from a freezer has a temperature of 35°F. Ten minutes after being placed on a table in a room with surrounding temperature 73°F, the ice cream\'s temperature is 0°F. Find the cooling constant k, correct to two decimal places, in Newton\'s Law of Cooling equation T(10) = 73 108e10k. (Enter a number.)

Solution

The Ice cream removed from a freezer has a temperature of 35°F. Ten minutes after being placed on a table in a room with surrounding temperature 73°F, the ice cream\'s temperature changes from -35⁰ F to 0°F. The Newton\'s Law of Cooling equation is T(t) = T_s+ (T₀ –T_s)e^-kt wheret is the time taken, T(t) is the temperature at time t, T_s is the surrounding temperature, T₀ is the initial temperature of the object and and k is the cooling constant. Here, t = 10 minutes, T_s= 73°F, T₀ =-35°F and t(10) = 0⁰F so that

0 = 73+ ( -35 -73)e^-10k or, -108 e-^10k = -73 or, e-^10k = 73/108 . Now, on taking natural log of both the sides, we get -10k ln e = ln (73/108) or -10k = ln 73 –ln 108 = 4.290459- 4.682131 = -0.391672. Hence k =0.391672/10 = 0.0391672 = 0.04( on rounding off to 2 decimal places)

Note:

ln e = 1 and ln(a/b) = ln a –ln b.