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For water in a particular boiler, the temperature $T$, in $^\\circ$C, as it cools down, is given by Newton's Law of Cooling:

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\\[T=\\simplify{{Ta}+{T0-Ta}e^({-k}*t)}\\]

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where $t$ is the time in seconds.

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Part a)

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{t1/60} minutes is {t1} seconds, so we substitute $t=\\var{t1}$ into the temperature formula:

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\\[T=\\var{Ta}+\\var{T0-Ta}e^{\\var{-k}\\times\\var{t1}}=\\var{precround(Temp1,4)}\\]

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which rounds to {precround(Temp1,1)}$^\\circ$C to 1 decimal place.

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Part b)

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Start by swapping sides:

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\\[\\simplify{{Ta}+{T0-Ta}e^({-k}*t)}=T\\]

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Subtract $\\var{Ta}$:

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\\[\\simplify{{T0-Ta}e^({-k}*t)}=T-\\var{Ta}\\]

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Divide by $\\var{T0-Ta}$:

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\\[\\simplify{e^({-k}*t)}=\\frac{T-\\var{Ta}}{\\var{T0-Ta}}\\]

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Take natural log:

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\\[\\simplify{{-k}*t}=\\ln\\left(\\frac{T-\\var{Ta}}{\\var{T0-Ta}}\\right)\\]

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And divide by $\\var{-k}$. A convenient way to do this, is to multiply by $-\\frac{1}{\\var{k}}$:

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\\[t=\\simplify{-1/{k}*ln((T-{Ta})/({T0}-{Ta}))}\\]

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Part c)

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Substitute $T=\\var{Temp2}$ into this formula:

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\\[t=-\\frac{1}{\\var{k}}\\times\\ln\\left(\\frac{\\var{Temp2}-\\var{Ta}}{\\var{T0-Ta}}\\right)=\\var{precround(t2,4)}\\]

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This rounds to {precround(t2,0)} seconds.

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Calculate the temperature after {t1/60} minutes, correct to 1 decimal place.

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$T=$ [[0]]$^\\circ$C

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Incorrect rounding/accuracy.

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Did not convert minutes into seconds.

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Select the correctly rearranged formula to make $t$ the subject.

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Hence calculate the time (to the nearest second) when the temperature reaches $\\var{Temp2}^\\circ$C.

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$t=$ [[0]] seconds

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Incorrect rounding/accuracy.

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