200 joules of heat flows into a 35.0 g sample. If the temperature increases by 10.0 °C, then what is the specific heat capacity of the sample, in J/kg °C?

Question

Quizplus · Accepted Answer

The specific heat capacity (c) can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat energy transferred (200 joules), $m$ is the mass of the substance (35.0 g or 0.035 kg), and $\Delta T$ is the change in temperature (10.0 °C). Rearranging the formula to solve for $c$, we get $c = \frac{q}{m\Delta T}$. Substituting the given values, $c = \frac{200}{0.035 \times 10} = \frac{200}{0.35} = 571.4 \, \text{J/kg °C}$, which rounds to 571 J/kg °C.

200 Joules of Heat Flows into a 35