Question

how many kilojoules of energy are required to bring 800.0 g water from 20.0°c to the boiling point?
268 kj
2.68×10^8 kj
3.35×10^5kj
3.35×10^8kj
question 8
6 pts
how many kilojoules of energy are released when.1.60 kg of water cools from 80.0 °c to 60.0 °c?
32.0 kj
382 kj
515 kj
134 kj

Explanation:

Step1: Identify the formula

The formula for heat energy is $q = mc\Delta T$, where $q$ is the heat energy, $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature. The specific - heat capacity of water $c = 4.184\ J/(g\cdot^{\circ}C)$.

Step2: Solve the first problem

For the first question, $m = 800.0\ g$, $\Delta T=T_{final}-T_{initial}=100.0^{\circ}C - 20.0^{\circ}C=80.0^{\circ}C$.
$q = mc\Delta T=(800.0\ g)\times(4.184\ J/(g\cdot^{\circ}C))\times80.0^{\circ}C$
$q = 800.0\times4.184\times80.0\ J=267776\ J$.
Convert to kJ: $q=\frac{267776\ J}{1000\ J/kJ}=267.776\ kJ\approx268\ kJ$.

Step3: Solve the second problem

First, convert the mass to grams. $m = 1.60\ kg=1600\ g$, $\Delta T=T_{final}-T_{initial}=60.0^{\circ}C - 80.0^{\circ}C=- 20.0^{\circ}C$.
$q = mc\Delta T=(1600\ g)\times(4.184\ J/(g\cdot^{\circ}C))\times(-20.0^{\circ}C)$
$q=1600\times4.184\times(- 20.0)\ J=-133888\ J$.
The negative sign indicates heat is released. Convert to kJ: $q=\frac{-133888\ J}{1000\ J/kJ}=-133.888\ kJ\approx - 134\ kJ$. The magnitude of the heat released is $134\ kJ$.

Answer:

First question: 268 kJ
Second question: 134 kJ