When a body is removed from an oven, its temperature is measured at 230∘F. Four minutes later its temperature is 120∘F. If the room temperature is found to be 70∘F, how long will it take for the chicken to cool off to a temperature of 90∘F. What is the steady state temperature of the body?

Name: When a body is removed from an oven, its temperature is measured at 230∘F. Four minutes later its temperature is 120∘F. If the room temperature is found to be 70∘F, how long will it take for the chicken to cool off to a temperature of 90∘F. What is the steady state temperature of the body?
Uploaded: 2022-02-23T17:22:57+00:00
Description: When a body is removed from an oven, its temperature is measured at 230∘F. Four minutes later its temperature is 120∘F. If the room temperature is found to be 70∘F, how long will it take for the chicken to cool off to a temperature of 90∘F. What is the steady state temperature of the body?

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lovagwb · Accepted Answer

According to question given that  When a body is removed from an oven measured temperature =23∘F  4 minutes later temperature is =120∘F  Room temperature =70∘F  Apply the Newtons

Mason Hall · Accepted Answer

T−Ts=(T0−Ts)e−kt  T0=210∘F  Ts=50∘F  T=100∘F  T−T=(T0−Ts)e−kt  100−50=(210−50)e−k×4  516=e−4k  ln⁡(516)=−4k  k=ln⁡(516)−4=0.291.  Find t hv t=70  70−50=(210−50)e−0.291×t  216=e−0.291t  ln⁡(18})=−0.291t  t=7.15min.

Vasquez · Accepted Answer

We have two constants to solve for, but if we plug in what we know we can find them. At t=0:210=Ce0+70⇒C=140 At t=30:140=140e−k⋅30+70⇒ek⋅30=12 ⇒k=−130ln⁡(12)=130ln⁡(2) So we now have the full function of T with respect to t T=140e−130ln⁡(2)t+70 Now to finish the problem we solve for t when T=100 100=140e−130ln⁡(2)t+7030140=e−130ln⁡(2)tln⁡(314)=−130ln⁡(2)tt=30ln⁡(143)ln⁡(2)t=30⋅2.22239=66.67177

When a body is removed from an oven, its temperature is meas

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