A 33.0 g of aluminum is heated on a stove. The temperature of the aluminum increases from 25 $\degree$C to 120 $\degree$C. How much heat energy was absorbed? (The specific heat of aluminum is 0.0215 cal/g$\degree$C.)

Question

Quizplus · Accepted Answer

The heat energy absorbed can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat absorbed, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Here, $m = 33.0 \,g$, $c = 0.0215 \,cal/g^\circ C$, and $\Delta T = 120^\circ C - 25^\circ C = 95^\circ C$. Thus, $q = 33.0 \,g \times 0.0215 \,cal/g^\circ C \times 95^\circ C = 67 \,cal$.

A 330 G of Aluminum Is Heated on a Stove °\degree°

A 330 G of Aluminum Is Heated on a Stove $\degree$