A 4.60-g sample of iron is heated from 47.0°C to 75.0°C. The amount of energy required is 57.96 J. The specific heat capacity of this sample of iron is
A) 0.450 J/g·°C
B) 1.233 J/g·°C
C) 2.07 J/g·°C
D) 1623 J/g·°C
E) 7465 J/g·°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 absorbed or released (57.96 J), $m$ is the mass of the substance (4.60 g), and $\Delta T$ is the change in temperature (75.0°C - 47.0°C = 28.0°C). Rearranging the formula to solve for $c$, we get $c = \frac{q}{m\Delta T}$. Plugging in the values, $c = \frac{57.96 \, \text{J}}{4.60 \, \text{g} \times 28.0 \, \text{°C}} = 0.450 \, \text{J/g·°C}$.

A 460-G Sample of Iron Is Heated from 47