Question

1. how much energy is released by 650g of water as it cools from its boiling point to its freezing point? the specific heat capacity of water is 4.18 j/g°c. enter your answer in the box provided. record your answer to the nearest kilojoules. kilojoules

Explanation:

Step1: Identify temperature - change

The boiling point of water is $100^{\circ}C$ and the freezing point is $0^{\circ}C$. So, $\Delta T=100 - 0=100^{\circ}C$.

Step2: Use the heat - transfer formula

The formula for heat transfer is $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Given $m = 650g$, $c=4.18J/g^{\circ}C$, and $\Delta T = 100^{\circ}C$. Then $Q=650\times4.18\times100$.
$Q = 650\times418=271700J$.

Step3: Convert to kilojoules

Since $1kJ = 1000J$, then $Q=\frac{271700}{1000}=271.7kJ$. Rounding to the nearest kilojoule, $Q\approx272kJ$.

Answer:

272