What is the specific heat of a $\pu{20 g}$ substance that releases $\pu{979 J}$ of heat when changing from $\pu{70 ^\circ C}$ to $\pu{20 ^\circ C}$?

I started out with the following equation: $c = \frac{Q}{m\Delta T}$ and $Q = 979\ \mathrm{J}$ and $m = 20\ \mathrm{g}$ and $\Delta T = 45\ \mathrm{^\circ C}$.

My final answer was $2202.75\ \mathrm{J}$

Did I use the right equations?


Your equation is correct, as are most of your measurements.

The unit for $c$ is $\ce{Jg^{-1}K^{-1}}$ if using grams as mass, ($\ce{Jkg^{-1}K^{-1}}$ if you convert the mass to kilograms).

Using your formula, however, does not give the answer you gave. When calculating using a calculator, consider that a parenthesis needs to be put around the denominator, as such:

$$c = \frac{q}{(m\Delta T)}$$

or else the calculator takes it as being $\ce{979J / 20g \times 45K}$ which gives the value you got.

It needs to be $\ce{979J / (20g \times 45K)}$ which gives an answer that is much more reasonable.

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