Suppose we place a $100-\mathrm{g}$ aluminum block with an initial temperature of $100^{\circ} \mathrm{C}$ in a Styrofoam cup containing a $100 \mathrm{~g}$ sample of water at $0^{\circ} \mathrm{C}$. (The specific heats of aluminum and water are $900 \mathrm{~J} \cdot \mathrm{kg}^{-1} \mathrm{~K}^{-1}$ and $4200 \mathrm{~J} \cdot \mathrm{kg}^{-1} \mathrm{~K}^{-1}$, respectively.) The system's final temperature is closest to
A) $0^{\circ} \mathrm{C}$.
B) $20^{\circ} \mathrm{C}$.
C) $50^{\circ} \mathrm{C}$.
D) $80^{\circ} \mathrm{C}$.
E) $100^{\circ} \mathrm{C}$.

Question

Quizplus · Accepted Answer

The heat lost by the aluminum block will be equal to the heat gained by the water. Using the formula $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature, we can set the heat lost by aluminum equal to the heat gained by water and solve for the final temperature. Given the significantly higher specific heat of water compared to aluminum, the final temperature will be closer to the initial temperature of water, but not exactly at $0^{\circ} \mathrm{C}$ due to the energy transfer, making $20^{\circ} \mathrm{C}$ the closest estimate.

Suppose We Place A 100−g100-\mathrm{g}100−g Aluminum Block with an Initial Temperature Of

Suppose We Place A $100-\mathrm{g}$ Aluminum Block with an Initial Temperature Of