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 we placed jello salad at 30 degrees Fahrenheit in a room with 68 degrees F, 5 mins later became 44 degrees F, what would be its temperature after 10 mins?

Final Answer:

T₃ = 52.8421 °F

Step-by-step explanation:

T_{0} = 30 °F T_{1} = 44 °F T_{2} = 68 °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 \frac{T _{2} - T _{1}}{T _{2} - T _{0}} k = ln (\frac{T _{2} - T _{1}}{T _{2} - T _{0}}) / t_{1} = ln (\frac{6 8 - 4 4}{6 8 - 3 0}) / 5 ≐ - 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 °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 °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 °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} = \frac{1 0 0 4}{1 9} °F = 52.8421 °F

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You need to know the following knowledge to solve this word math problem:

algebra

basic operations and concepts

Units of physical quantities

temperature

themes, topics

fundamentals of Physics

Grade of the word problem

high school

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