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:

T0=30 F T1=44 F T2=68 F  t1=5 min  T1=T2(T2T0) ek t1  ek t1=(T2T1)/(T2T0) k t1=lnT2T1T2T0  k=ln(T2T1T2T0)/t1=ln(68446830)/50.0919  t0=0 min T0=T2(T2T0) ek t0=68(6830) e(0.0919) 0=30 F  t5=5 min T5=T2(T2T0) ek t5=68(6830) e(0.0919) 5=44 F  t8=1000 min T8=T2(T2T0) ek t8=68(6830) e(0.0919) 1000=68 F  t3=10 min T3=T2(T2T0) ek t3=68(6830) e(0.0919) 10=10041952.8421=52.842FT_{ 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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