A solid was heated and left to cool in a container in a room. The difference between its temperature and the surrounding room temperature at time $t$ hours was $T^\circ\text{C}$. The table shows some recorded values of $t$ and $T$. The variables $t$ and $T$ are related by the equation $T = a e^{b t}$, where $a$ and $b$ are constants.

1. Calculate the values of $\ln T$ corresponding to each given value of $t$.

Taking the natural logarithm of the given $T$ values (to 2 decimal places):

2. Choose suitable scales, plot the points $(t, \ln T)$ on the grid, and draw the line of best fit.

1. Set the horizontal axis for $t$ from $0$ to $25$ (or $30$).
2. Set the vertical axis for $\ln T$ from $0$ to $3.0$.
3. Plot the points $(5, 2.71)$, $(10, 2.13)$, $(15, 1.65)$, $(22, 0.99)$, and $(25, 0.47)$.
4. Draw a straight line of best fit passing through these points.

1. Express the given equation in linear form

$T = a e^{bt}$
Taking the natural logarithm on both sides:
$\ln T = \ln(a e^{bt})$
$\ln T = bt + \ln a$
This is in the form $Y = mX + c$, where $Y = \ln T$, $X = t$, gradient $m = b$, and $Y$-intercept $c = \ln a$.

2. Calculate the gradient to find $b$

Using the points $(5, \ln 15.1)$ and $(25, \ln 1.6)$ from the line of best fit:
$b = \frac{\ln 1.6 - \ln 15.1}{25 - 5}$
$b = \frac{0.4700 - 2.7147}{20}$
$b \approx -0.112$

3. Find the $Y$-intercept to calculate $a$

Substitute $t = 5$, $\ln T = 2.7147$, and $b = -0.1122$ into the linear equation:
$2.7147 = -0.1122(5) + \ln a$
$\ln a = 2.7147 + 0.5610 = 3.2757$
$a = e^{3.2757} \approx 26.5$

1. Explain why $T$ cannot be zero with reference to the graph of $\ln T$

The graph in (a) plots $\ln T$ against $t$. If $T = 0$, we would need to evaluate $\ln 0$, which is undefined.