Calculate the heat released when 10.0 g of water at 25.0 °C cools to ice at 0.0 °C.The specific heat of water is 1.00 cal/(g x °C);the heat of fusion is 80.0 cal/g;and the heat of vaporization is 540.0 cal/g.

Question

Quizplus · Accepted Answer

First, calculate the heat released when the water cools from 25.0 °C to 0.0 °C using the formula $q = mc\Delta T$, where $m = 10.0 \, \text{g}$, $c = 1.00 \, \text{cal/(g°C)}$, and $\Delta T = 25.0 \, \text{°C}$. This gives $q = 10.0 \, \text{g} \times 1.00 \, \text{cal/(g°C)} \times 25.0 \, \text{°C} = 250.0 \, \text{cal}$.Then, calculate the heat released during the phase change from water at 0.0 °C to ice at 0.0 °C using the heat of fusion, $q = m\Delta H_{\text{fusion}}$, where $\Delta H_{\text{fusion}} = 80.0 \, \text{cal/g}$ and $m = 10.0 \, \text{g}$. This gives $q = 10.0 \, \text{g} \times 80.0 \, \text{cal/g} = 800.0 \, \text{cal}$.The total heat released is the sum of these two values: $250.0 \, \text{cal} + 800.0 \, \text{cal} = 1050.0 \, \text{cal}$, or $1.05 \times 10^3 \, \text{cal}$.

Calculate the Heat Released When 10