The temperature T of a cooling object drops at a rate proportional to the difference `T - S` where S is constant temperature of surrounding medium. If initially `T = 150^@C`; find the temperature of the cooling object at any time t.

The temperature T of a cooling object drops at a rate proportional to the difference (T-s) where S is a constant temperature of surrounding medium. If initially T = 150^(@ C), find the temperature of the cooling object at any time t.

The temperature T of a cooling object drops at a rate proportional to the difference T-S, where S is the constant temperature of the surrounding medium. Thus, (dT)/(dt)=-k(T-S) , where k(>0) is constant and t is the time. Solve the differntial equation if it is given that T(0)=150.

The temperature T of a cooling drops at a rate proportional to the difference T-S, where S is the constant temperature of surrounding medium. A thermometer reading 80^@F was taken outside. Five minutes later, the thermometer read 60^@F . After another five minutes the reading was 50^@F . what was the outside temperature?

Newton's law of cooling states that the rate of change of the temperature T of an object is proportional to the difference between T and the (constant) temperature tau of the surrounding medium, we can write it as (dT)/(dt) = -k(T - tau) k gt 0 constant An cup of coffee is served at 185^(@)F in a room where the temperature is 65^(@)F . 2 minutes later the temperature of the coffee has dropped to 155^(@)F . log_(e)3 = 1.09872, log_(e).(3)/(4) = 0.2877 Temperature of coffee at time t is given by

Newton's law of cooling states that the rate of change of the temperature T of an object is proportional to the difference between T and the (constant) temperature tau of the surrounding medium, we can write it as (dT)/(dt) = -k(T - tau) k gt 0 constant An cup of coffee is served at 185^(@)F in a room where the temperature is 65^(@)F . 2 minutes later the temperature of the coffee has dropped to 155^(@)F . log_(e)3 = 1.09872, log_(e).(3)/(4) = 0.2877 Time required for coffee to have 105^(@)F temperature is

As difference of temperature of body and surroundings increases, rate of cooling