A 22.0 g block of copper at 45^$\circ$C absorbs 2.50 kJ of heat. Given the specific heat of Cu is 0.385 J/g·^$\circ$C what will be the final temperature of the Cu

Question

Quizplus · Accepted Answer

To find the final temperature of the copper block, we use the formula for heat transfer: $q = mc\Delta T$, where $q$ is the heat absorbed, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Rearranging for $\Delta T$, we get $\Delta T = \frac{q}{mc}$. Substituting the given values: $\Delta T = \frac{2500\, \text{J}}{22.0\, \text{g} \times 0.385\, \text{J/g}\,^\circ\text{C}} = \frac{2500}{8.47} \approx 295.16\,^\circ\text{C}$. Adding this to the initial temperature of $25\,^\circ\text{C}$ (since $45\,^\circ\text{F}$ is approximately $7.22\,^\circ\text{C}$, and there seems to be a formatting error), gives a final temperature significantly higher than any of the options provided. However, assuming the initial temperature was meant to be $25\,^\circ\text{C}$ and recognizing a possible error in the interpretation due to formatting, the closest correct option based on the calculation method would be $340\,^\circ\text{C}$, considering potential errors in interpreting the initial temperature and the options provided.

A 220 G Block of Copper at 45 ∘\circ∘ C Absorbs 2

A 220 G Block of Copper at 45^$\circ$C Absorbs 2