I am doing an online assignment where I am being asked to find temperature change but there is something wrong in my calculations as I am getting the answer wrong. Can someone help me figure out where I went wrong in my calculations?

Assignment Instructions:

Solve for the unknown variable using the specific heat formula. Input your answer (without units) and using the correct number of sig figs.


If 31,500J of heat is used to warm 750g of water, what is the temperature change? The specific heat of water is 4.184J/goC.

My Answer: 10

Since I am being asked to find the change in temperature in this question, I split the original equation up. First, I did (750) * (4.184) and got 3138. Then, I did (31,500) / (3138) and got 10.0382409178. The lowest number of significant figures is two, so I rounded 10.0382409178 to 10. Why is the computer counting my question as wrong?

I used the equation temperature change = (heat) / (mass) * (specific heat). The variables I used are q (heat), m (mass), and c (specific heat).

  • $\begingroup$ See my answer below. This is bad software, and our new visitor is a victim of it. $\endgroup$
    – Todd Minehardt
    Mar 27 at 20:32

You're not really wrong: the computer program is apparently expecting a float and not an integer, which is really stupid because the question is about temperature change and not the vagaries of significant figures.

Instead of you learning about heat capacity and the relation to temperature change, you're instead learning about significant figures first, and the chemistry/physics part second.

I find this sort of thing patently awful and a frustrating way for you or anyone else to "learn" anything about the principles.

You have to report your answer to 2 significant figures: 750 has 2 (the limiting number); 31500 has 3; and 4.184 has 4.

$$Q = m\times c\times \Delta T$$


$$\Delta T = {Q\over m\times c}$$


$$\Delta T = {31,500\ {\rm J}\over 750\ {\rm g} \times 4.184\ {\rm J}\cdot {\rm g}^{-1}\cdot {\rm ^{\circ}C}^{-1}}$$

which, to 2 significant figures, is:

$$\Delta T = 10.\ ^{\circ}{\rm C}$$

Notice I have included a decimal point, so the now ridiculous answer that should be accepted is 10. and not 10

Meanwhile, you've lost a lot of time wondering where your calculation went wrong because some programmer doesn't know how to cast. Or maybe you're really supposed to report numbers with a terminal decimal point, and I'll challenge anyone to find any respected publication where that happens:

"The temperature change, in degrees Celsius, is 10.."

Note that 10.0 has three significant figures, so that's also "wrong."

And that, my friend, is why "learning" from a computer is a lazy way out for the "teacher."

  • $\begingroup$ Thank you, Todd. I understand where I went wrong now! $\endgroup$ Mar 27 at 20:53

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