# Exponential warm

Suppose that a body with temperature T1 is placed in surroundings with temperature T0 different from that of T1. The body will either cool or warm to temperature T(t) after time t, in minutes, where T(t)=T0 + (T1-T0)e^(-kt) .

If jello salad with 30 degree F placed in a room with 68 degree F, 5 mins later became 44 degree F, what will be its temperature after 10 mins?

Result

T3 =  52.842 °F

#### Solution:

$T_{ 0 } = 30 \ ^\circ F \ \\ T_{ 1 } = 44 \ ^\circ F \ \\ T_{ 2 } = 68 \ ^\circ F \ \\ \ \\ t_{ 1 } = 5 \ min \ \\ \ \\ T_{ 1 } = T_{ 2 } - (T_{ 2 }-T_{ 0 }) \cdot \ e^{ k \cdot \ t_{ 1 } } \ \\ \ \\ e^{ k \cdot \ t_{ 1 } } = (T_{ 2 }-T_{ 1 })/(T_{ 2 }-T_{ 0 }) \ \\ k \cdot \ t_{ 1 } = \ln \dfrac{ T_{ 2 }-T_{ 1 } }{ T_{ 2 }-T_{ 0 } } \ \\ \ \\ k = \ln (\dfrac{ T_{ 2 }-T_{ 1 } }{ T_{ 2 }-T_{ 0 } } ) / t_{ 1 } = \ln (\dfrac{ 68-44 }{ 68-30 } ) / 5 \doteq -0.0919 \ \\ \ \\ t_{ 0 } = 0 \ min \ \\ T_{ 0 } = T_{ 2 } - (T_{ 2 }-T_{ 0 }) \cdot \ e^{ k \cdot \ t_{ 0 } } = 68 - (68-30) \cdot \ e^{ (-0.0919) \cdot \ 0 } = 30 \ ^\circ F \ \\ \ \\ t_{ 5 } = 5 \ min \ \\ T_{ 5 } = T_{ 2 } - (T_{ 2 }-T_{ 0 }) \cdot \ e^{ k \cdot \ t_{ 5 } } = 68 - (68-30) \cdot \ e^{ (-0.0919) \cdot \ 5 } = 44 \ ^\circ F \ \\ \ \\ t_{ 8 } = 1000 \ min \ \\ T_{ 8 } = T_{ 2 } - (T_{ 2 }-T_{ 0 }) \cdot \ e^{ k \cdot \ t_{ 8 } } = 68 - (68-30) \cdot \ e^{ (-0.0919) \cdot \ 1000 } = 68 \ ^\circ F \ \\ \ \\ t_{ 3 } = 10 \ min \ \\ T_{ 3 } = T_{ 2 } - (T_{ 2 }-T_{ 0 }) \cdot \ e^{ k \cdot \ t_{ 3 } } = 68 - (68-30) \cdot \ e^{ (-0.0919) \cdot \ 10 } = \dfrac{ 1004 }{ 19 } \doteq 52.8421 = 52.842 ^\circ F$

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