According to Newton's law of cooling, the rate of cooling dT/dt = k(T - R) when T is the temperature of the object and R is the ambient temperature, k is a constant. The differential equation can be solved to give: T(t) = Ta + (To - Ta)e^-(k*t)

Here the constant k = 0.2231 per minute.

The original difference in temperature To - Ta = 50. The ambient temperature Ta = 80. We have to find t at which T = 82 degrees.

T(t) = 82 = 80 + 50*e^-(0.2231*t)

=> 2 = 50*e^-(0.2231*t)

=> e^(0.2231*t) = 25

take the log of both the sides

=> 0.2231*t = ln 25 = 3.2188

=> t = 14.42

**The temperature reduces from 130 to 82 in 14.42 min.**