dT/dt = -k(T-A)
dT/(T-A) = -k dt
ln(T-A) = -kt + c
T-A = c e^(-kt)
T = A + c e^(-kt)
Now just plug in you data point, and you have
30 + c e^-k = 80
ce^-k = 50
Now, what you left out is that c = 100-30 = 70
e^-k = 5/7
k = ln(7/5) = 0.336
So, now you know that
T(t) = 30 + 70e^(-1.25t)
So now you can find T(4)
The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT/dt = -k(T - A), where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 30°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.
4 answers
T(t) = 30 + 70e^(-1.25t) isn't correct answer.
It's like this.
k = ln ( 7 / 5 )
T = A + c ∙ e ^ ( - k t )
T = 30 + 70 ∙ e ^ ( - k t )
T = 30 + 70 ∙ e ^ [ - ln ( 7 / 5 ) ∙ t ]
T = 30 + 70 ∙ [ e ^ ln ( 7 / 5 ) ] ^ ( - t )
T = 30 + 70 ∙ ( 7 / 5 ) ^ ( - t )
T = 30 + 70 ∙ 1.4 ^ ( - t )
T(4) = 30 + 70 ∙ 1.4 ^ ( - 4 )
T(4) = 30 + 70 ∙ 0.26030825
T(4) = 30 + 18.2215775
T(4) = 48.2215775 °C
T(4) = 48°C
to the nearest °C
It's like this.
k = ln ( 7 / 5 )
T = A + c ∙ e ^ ( - k t )
T = 30 + 70 ∙ e ^ ( - k t )
T = 30 + 70 ∙ e ^ [ - ln ( 7 / 5 ) ∙ t ]
T = 30 + 70 ∙ [ e ^ ln ( 7 / 5 ) ] ^ ( - t )
T = 30 + 70 ∙ ( 7 / 5 ) ^ ( - t )
T = 30 + 70 ∙ 1.4 ^ ( - t )
T(4) = 30 + 70 ∙ 1.4 ^ ( - 4 )
T(4) = 30 + 70 ∙ 0.26030825
T(4) = 30 + 18.2215775
T(4) = 48.2215775 °C
T(4) = 48°C
to the nearest °C
It is better to keep k value in exact form till you get final answer. So use k=ln(5/7) to find T(4).
Also ln(5/7) can be written as -ln(7/5) using properties of logs
T(4)=30+70e^(ln(7/5)*4) =48.2
Also ln(5/7) can be written as -ln(7/5) using properties of logs
T(4)=30+70e^(ln(7/5)*4) =48.2
The temperature, f(t) of a cup of coffee after t minutes can be determined by the equation f(t)=63(0.91)T