Question

Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. So the rate of cooling for a bottle of lemonade at a room temperature of 75°F which is placed into a refrigerator with temperature of 38°F can be modeled by dT dt equals k times the quantity T minus 38 where T(t) is the temperature of the lemonade after t minutes and T(0) = 75. After 30 minutes the lemonade has cooled to 60°F, so T(30) = 60.
To the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?

To the nearest minute, how long does it take for the lemonade to cool to 55°F?

Answer 1

a) The initial temperature difference of 75-38 = 37 degrees has been reduced to 60-38 = 22 degrees in 30 minutes. According to the law of cooling, it will be reduced by the same factor of 22/37 in another 30 minutes, so the temperature difference then will be 22(22/37) = 13 degrees.

The temperature of the lemonade after an additional 30 minutes will be 38+13= 51°F.

b) The temperature can be modeled by
  T(t) = 38 + 37·(22/37)^(t/30)
Substituting the given information, we have
  T(t) = 55 = 38 +37·(22/37)^(t/30)
  17/37 = (22/37)^(t/30) . . . . . . subtract 38, divide by 37
  log(17/37) = (t/30)·log(22/37) . . . . take the logarithm
  t = 30·log(17/37)/log(22/37) . . . . divide by the coefficient of t
  t ≈ 44.878

It will take about 45 minutes for the lemonade to cool to 55 °F.