Solution:
Use Newton's Law of Cooling.
T = T_s + (T_0 - T_s)*e^(-kt)
where
T = temperature at any instant
T_s = temperature of surroundings
T_0 = original temperature
t = elapsed time
k = constant
Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.
T = 14, T_0 = 24, T_s = -15, t = 1, k = ?
T = T_s + (T_0 - T_s)*e^(-kt)
==> 14 = -15 + (24 - 7)*e^(-k)
==> 14 = 7 + 17*e^(-k)
==> 7 = 17*e^(-k)
==> 7/13 = e^(-k)
==> -k = ln(7/17)
==> k = -ln(7/17) ≈ 0.774
Now,
Let's calculate temperatures!
T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3
T = T_s + (T_0 - T_s)*e^(-kt)
==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]
==> T = -15 + 39*e^(-1.548)
==> T ≈ 15.72° C
This the required answer.
