200 joules of heat flows into a $35.0 \mathrm{~g}$ sample. If the temperature increases by $10.0^{\circ} \mathrm{C}$, then what is the specific heat capacity of the sample, in $\mathrm{J} / \mathrm{kg}^{\circ} \mathrm{C}$ ?

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 ($200 \, \text{J}$), $m$ is the mass of the sample ($35.0 \, \text{g} = 0.035 \, \text{kg}$), and $\Delta T$ is the change in temperature ($10.0^{\circ} \text{C}$). Rearranging the formula to solve for $c$, we get $c = \frac{q}{m\Delta T}$. Substituting the given values, $c = \frac{200 \, \text{J}}{0.035 \, \text{kg} \times 10.0^{\circ} \text{C}} = 571.4 \, \text{J/kg}^{\circ} \text{C}$, which rounds to $571 \, \text{J/kg}^{\circ} \text{C}$.

200 Joules of Heat Flows into A 35.0 g35.0 \mathrm{~g}35.0 g Sample If the Temperature Increases By A 35.0

200 Joules of Heat Flows into A $35.0 \mathrm{~g}$ Sample If the Temperature Increases By A 35.0