Assume that 491.8 J of heat is added to 5.00 g of water originally at 23.0°C. What would be the final temperature of the water? (Specific heat capacity of water = 4.184 J/g·°C.)

Question

Quizplus · Accepted Answer

The final temperature can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat added, $m$ is the mass of the water, $c$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature. Rearranging the formula to solve for $\Delta T$, we get $\Delta T = \frac{q}{mc}$. Plugging in the given values, $\Delta T = \frac{491.8\, \text{J}}{5.00\, \text{g} \times 4.184\, \text{J/g}·°\text{C}} = 23.5\, °\text{C}$. Adding this temperature change to the original temperature of 23.0°C gives a final temperature of 46.5°C.

Assume That 491