Am I Using the Correct Number of Significant Figures (Heat Problem)?
This is sort of tricky, but I'd say that neither 10.8 nor 11 is right! 

You got
$$Q_1 = 9.7\bar22753346\ \mathrm{kcal}$$ 

which you rounded to $9.72\ \mathrm{ kcal}$. The bar over the first two indicates that is the significant figure and that the excess numbers can be carried in intermediate calculations but would be trimmed for the answer.

Now for the second part you got

$$Q_2 = 1.\bar134\ \mathrm{ kcal}$$ 

which you rounded to  $1.1\ \mathrm{ kcal}$. 

Using the intermediate calculations you are calculating the sum. 

$$\begin{align}
9&amp;.7\bar22753346\\
1&amp;.\bar134\\
Q_1+Q_2 =10&amp;.\bar856753346 \end{align}$$

Since 9.72+ answer is good to hundredths, but 1.1134 answer is only good to the tenths, thus the answer must be rounded to the tenths. So the "best" answer using significant figures is:
$$Q_1+Q_2 = 10.9\ \mathrm{ kcal}$$ 

You should not round intermediate results to their "correct" number of significant figures since that leads to rounding errors in the intermediate calculations.  

$$\begin{align}
9&amp;.7\bar2\\
1&amp;.\bar1\\
Q_1+Q_2 =10&amp;.\bar82 \end{align}$$

Obviously carrying the ten digits in $9.7\bar22753346$ is somewhat ridiculous, so a good rule of thumb with a calculator would be to carry two extra digits in intermediate calculations. So round the intermediate number to $9.7\bar228$. The value $1.\bar134$ has two extra digits so its fine as shown. 

Notice that this is exactly the same rule you used in calculating the temperature difference.

$$\begin{align}
10\bar0&amp;.\\
-37&amp;.\bar0\\
=6\bar3&amp;.\end{align}$$

Had it been 37.5 degrees for body temperature, then you should have used $6\bar2.5$ in the intermediate calculations. 

$$\begin{align}
10\bar0&amp;.\\
-37&amp;.\bar5\\
=6\bar2&amp;.5\end{align}$$

As a matter of curiosity I used the rnd() function in Excel to simulate uniform distribution between all the values. So:


17.95 grams to 18.04999 grams of water
99.5 to 100.49999 degrees C
36.95 to 37.04999 degrees C


$\text{=(2.26)*(18+0.1*(RAND()-0.5))/4.184+(18+0.1*(RAND()-0.5))*(100+(RAND()-0.5)-(37+0.1*(RAND()-0.5)))/1000}$

The 4.184 and 2260 values were assume to be exact. Averaging over a thousand such values I got 10.85703268
with a standard deviation of 0.016370732. 

So the gist is that using significant figures doesn't give as good a result as possible. The problem is in the use of 2 significant figures in the calculation of $Q_2$. For the difference between the uniform differences of 100 and 37.0 the relative standard deviation is 0.46%. However the uniform distribution of 1.05 to 1.15 has a relative standard deviation of 2.6%, much worse than what is really happening.