The question is:

A patient arrives in the emergency room with a burn caused by steam. Calculate the heat, in kilocalories, that is released when $\pu{18.0 g}$ of steam at $\pu{100 °C}$ hits the skin, condenses, and cools to body temperature of $\pu{37.0 °C}$.

1) I solve for the steam as it condenses:

Heat of vaporization: $$\Delta H_\text{vap}(\ce{H2O}) = \pu{2260 J/g}$$

Heat: $$\begin{align}Q_1 &= \pu{18.0 g} \times (\pu{2260 J/g}) \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/ 10^3 cal})\\ &= \pu{9.722753346 kcal} \to \text{rounded to 3 SF} = \pu{9.72 kcal}\end{align}$$

2) I solve for the cooling of the $\ce{H2O}$:

Specific heat capacity: $$c(\ce{H2O}) = 4.184\ \mathrm{J/(g\ ^\circ C)}$$

Temperature change: $$\Delta T = \pu{100 °C} - \pu{37.0 °C} = \pu{63 °C}$$ (2 SF)

Heat: $$\begin{align}Q_2&=\pu{18.0 g} \times \pu{63 °C} \times 4.184\ \mathrm{J/(g\ ^\circ C)} \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/10^3 cal})\\&= \pu{1.134 kcal} \to \text{rounded to 2 SF} = \pu{1.1 kcal}\end{align}$$

I then add the two results together and get:

$$\begin{align}Q&=Q_1+Q_2\\&=\pu{9.72 kcal} + \pu{1.1 kcal} = \pu{10.82 kcal}\end{align}$$

$\to \text{rounded to 3 SF} = \pu{10.8 kcal}$ (using the addition/subtraction rule for SFs)

The textbook's answer is $\pu{11 kcal}$.

Is my answer right, according to significant figure rules, or is the textbook's provided answer correct?

  • Todd, I have a quick question: in an equation, such as specific heat, is the temperature change (Delta T) irrelevant as a significant figure? It's causing me a ton of grief. And, do you know of any texts or resources that may help me understand multi-step problems, in regards to significant figure amounts? Thanks a lot. – Don't Think. Feel. Dec 23 '17 at 3:00
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    Actually 18.0 has 3 significant figures. 18 would have two – Raven Dec 23 '17 at 19:15
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    Nice job of clearly showing all your work so someone could find your mistake and give an answer. That is what we ask of homework type of problems, so I have no idea why the downvotes. Also, welcome to Chemistry.SE. – airhuff Dec 23 '17 at 20:22
  • @Raven - Thanks for the catch. Ignore my previous comments. – Todd Minehardt Dec 23 '17 at 22:02

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.

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

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.

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

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.

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

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

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

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.

  • 1
    Thanks for all the responses! I had just delved deeper into this problem as well, and after deciding that I didn't think 11 kJ was the correct answer, I was also able to find that 10.9 kJ would be the most "precise" answer. I have to say, this problem really helped me to develop stronger SF skills! I'm taking CHEM 101 next semester, so I'm just trying to be as prepared as I can. – Don't Think. Feel. Dec 24 '17 at 2:04
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    I'd wonder if any chemistry book used SF correctly though out the book. So the book would have problems like "5 grams of X". Is that really just 5 grams, or 5.0 grams, or 5.00 grams etc... back in the day I used a slide rule to solve the problems. 3 SF was pretty much it, even on intermediate calculations. Rounding errors are the bane of calculations. You can't totally avoid the problem, you can just try to minimize it. – MaxW Dec 24 '17 at 2:31
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    Dually noted, MaxW! Thanks for your help. This definitely taught me not to round my intermediates, and implement guard digits (like 2 more SF than needed), while doing intermediate calculations. It's amazing how it affects the end result. I joined this website now, so maybe I can contribute a bit, myself. – Don't Think. Feel. Dec 24 '17 at 2:37

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