Question

over a period of time, a hot object cools to the temperature of the surrounding air. this is described mathematically by newtons law of cooling ( t = c+(t_0 - c)e^{-kt} ), where ( t ) is the time it takes for an object to cool from temperature ( t_0 ) to temperature ( t ), ( c ) is the surrounding air temperature, and ( k ) is a positive constant that is associated with the cooling object. a cake removed from the oven has a temperature of ( 213^{circ}f ) and is left to cool in a room that has a temperature of ( 71^{circ}f ). after 20 minutes, the temperature of the cake is ( 142^{circ}f ). what is the temperature of the cake after 35 minutes? after 35 minutes, the temperature of the cake is ( square^{circ}f ). (do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Identify the values

We know that \(C = 71\), \(T_0=213\). When \(t = 20\), \(T = 142\). First, substitute these values into the Newton's Law of Cooling formula \(T = C+(T_0 - C)e^{-kt}\).
So, \(142=71+(213 - 71)e^{-20k}\).

Step2: Solve for \(k\)

Subtract 71 from both sides: \(142 - 71=(213 - 71)e^{-20k}\).
\(71 = 142e^{-20k}\).
Divide both sides by 142: \(\frac{71}{142}=e^{-20k}\).
Since \(\frac{71}{142}=\frac{1}{2}\), we have \(\frac{1}{2}=e^{-20k}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{-20k})\).
Using the property \(\ln(e^x)=x\), we get \(\ln(\frac{1}{2})=- 20k\).
Since \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\), then \(-0.693=-20k\).
Solve for \(k\): \(k=\frac{0.693}{20}=0.03465\).

Step3: Find the temperature when \(t = 35\)

Now that we have \(k = 0.03465\), use the formula \(T = C+(T_0 - C)e^{-kt}\) with \(t = 35\), \(C = 71\), \(T_0=213\).
\(T=71+(213 - 71)e^{-0.03465\times35}\).
First, calculate the exponent: \(-0.03465\times35=-1.21275\).
\(e^{-1.21275}\approx0.3\).
Then \(T = 71+142\times0.3\).
\(T=71 + 42.6\).
\(T=113.6\).

Answer:

\(114\)