300.0g of water is at 12.0^ºC.The water absorbs 4016 J of heat energy.What is the new temperature of the water?

Question

Quizplus · Accepted Answer

The specific heat capacity of water is $4.18 \, \text{J/g}^\circ\text{C}$. To find the temperature change, use the formula $\Delta Q = mc\Delta T$, where $\Delta Q = 4016 \, \text{J}$, $m = 300.0 \, \text{g}$, and $c = 4.18 \, \text{J/g}^\circ\text{C}$. Solving for $\Delta T$ gives $\Delta T = \frac{4016 \, \text{J}}{300.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C}} = 3.2^\circ\text{C}$. Adding this to the initial temperature of $12.0^\circ\text{C}$ gives a new temperature of $15.2^\circ\text{C}$.

3000g of Water Is at 12