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Sagot :
Let's solve the problem step by step:
i) Find the temperature of the room.
We are given:
- Initial temperature of the body, [tex]\( T_{\text{initial}} = 40 \)[/tex]°C
- Temperature after 5.0 minutes, [tex]\( T_{\text{final}} = 30 \)[/tex]°C
- Cooling constant, [tex]\( k = 0.02 \)[/tex] W/m²K
- Initial time, [tex]\( t_{\text{initial}} = 5.0 \)[/tex] minutes
Using Newton's Law of Cooling, the temperature of the body as a function of time is given by:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
From the formula, after 5.0 minutes:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot e^{-0.02 \cdot 5} \][/tex]
Simplifying the equation, we have:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot e^{-0.1} \][/tex]
Let [tex]\( e^{-0.1} \approx 0.9048 \)[/tex]:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot 0.9048 \][/tex]
[tex]\[ 30 = T_{\text{room}} + 36.192 - 0.9048 T_{\text{room}} \][/tex]
Combining like terms:
[tex]\[ 30 = 36.192 - 0.9048 T_{\text{room}} + T_{\text{room}} \][/tex]
[tex]\[ 30 = 36.192 + 0.0952 T_{\text{room}} \][/tex]
Rearranging to solve for [tex]\( T_{\text{room}} \)[/tex]:
[tex]\[ 0.0952 T_{\text{room}} = 30 - 36.192 \][/tex]
[tex]\[ 0.0952 T_{\text{room}} = -6.192 \][/tex]
[tex]\[ T_{\text{room}} = \frac{-6.192}{0.0952} \][/tex]
[tex]\[ T_{\text{room}} = 485.41659723875216 \text{°C} \][/tex]
Thus, the temperature of the room is approximately 485.42°C.
ii) What will be the temperature of the body after 15.0 minutes from the initial process?
We need to find the temperature of the body, [tex]\( T(15) \)[/tex], after 15.0 minutes.
We use the same cooling formula:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
Plugging in the values for [tex]\( t = 15 \)[/tex] minutes:
[tex]\[ T(15) = 485.41659723875216 + (40 - 485.41659723875216) \cdot e^{-0.02 \cdot 15} \][/tex]
Simplifying:
[tex]\[ T(15) = 485.41659723875216 + (-445.41659723875216) \cdot e^{-0.3} \][/tex]
Let [tex]\( e^{-0.3} \approx 0.7408 \)[/tex]:
[tex]\[ T(15) = 485.41659723875216 + (-445.41659723875216) \cdot 0.7408 \][/tex]
[tex]\[ T(15) = 485.41659723875216 - 329.97273102851773 \][/tex]
[tex]\[ T(15) = 155.44386621023443 \text{°C} \][/tex]
Thus, the temperature of the body after 15.0 minutes from the initial process is approximately 155.44°C.
i) Find the temperature of the room.
We are given:
- Initial temperature of the body, [tex]\( T_{\text{initial}} = 40 \)[/tex]°C
- Temperature after 5.0 minutes, [tex]\( T_{\text{final}} = 30 \)[/tex]°C
- Cooling constant, [tex]\( k = 0.02 \)[/tex] W/m²K
- Initial time, [tex]\( t_{\text{initial}} = 5.0 \)[/tex] minutes
Using Newton's Law of Cooling, the temperature of the body as a function of time is given by:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
From the formula, after 5.0 minutes:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot e^{-0.02 \cdot 5} \][/tex]
Simplifying the equation, we have:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot e^{-0.1} \][/tex]
Let [tex]\( e^{-0.1} \approx 0.9048 \)[/tex]:
[tex]\[ 30 = T_{\text{room}} + (40 - T_{\text{room}}) \cdot 0.9048 \][/tex]
[tex]\[ 30 = T_{\text{room}} + 36.192 - 0.9048 T_{\text{room}} \][/tex]
Combining like terms:
[tex]\[ 30 = 36.192 - 0.9048 T_{\text{room}} + T_{\text{room}} \][/tex]
[tex]\[ 30 = 36.192 + 0.0952 T_{\text{room}} \][/tex]
Rearranging to solve for [tex]\( T_{\text{room}} \)[/tex]:
[tex]\[ 0.0952 T_{\text{room}} = 30 - 36.192 \][/tex]
[tex]\[ 0.0952 T_{\text{room}} = -6.192 \][/tex]
[tex]\[ T_{\text{room}} = \frac{-6.192}{0.0952} \][/tex]
[tex]\[ T_{\text{room}} = 485.41659723875216 \text{°C} \][/tex]
Thus, the temperature of the room is approximately 485.42°C.
ii) What will be the temperature of the body after 15.0 minutes from the initial process?
We need to find the temperature of the body, [tex]\( T(15) \)[/tex], after 15.0 minutes.
We use the same cooling formula:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
Plugging in the values for [tex]\( t = 15 \)[/tex] minutes:
[tex]\[ T(15) = 485.41659723875216 + (40 - 485.41659723875216) \cdot e^{-0.02 \cdot 15} \][/tex]
Simplifying:
[tex]\[ T(15) = 485.41659723875216 + (-445.41659723875216) \cdot e^{-0.3} \][/tex]
Let [tex]\( e^{-0.3} \approx 0.7408 \)[/tex]:
[tex]\[ T(15) = 485.41659723875216 + (-445.41659723875216) \cdot 0.7408 \][/tex]
[tex]\[ T(15) = 485.41659723875216 - 329.97273102851773 \][/tex]
[tex]\[ T(15) = 155.44386621023443 \text{°C} \][/tex]
Thus, the temperature of the body after 15.0 minutes from the initial process is approximately 155.44°C.
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