Question

how many joules of energy are released when 45.0 g of water cools from 18.0 °c to 7.0 °c?
495 j
791 j
2860 j
2070 j
question 6
6 pts
how many joules of energy are released when 100.0 g of water cools from 98.0 °c to 20.0 °c?
32600 j
32800 j
167 j
7800 j

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 $c = 4.18\ J/(g\cdot^{\circ}C)$.

Step2: Calculate $\Delta T$ for the first question

$\Delta T_1=18.0^{\circ}C - 7.0^{\circ}C=11.0^{\circ}C$, $m_1 = 45.0\ g$. Then $q_1=m_1c\Delta T_1=45.0\ g\times4.18\ J/(g\cdot^{\circ}C)\times11.0^{\circ}C = 45.0\times4.18\times11.0\ J=2070.9\ J\approx2070\ J$.

Step3: Calculate $\Delta T$ for the second question

$\Delta T_2=98.0^{\circ}C - 20.0^{\circ}C = 78.0^{\circ}C$, $m_2 = 100.0\ g$. Then $q_2=m_2c\Delta T_2=100.0\ g\times4.18\ J/(g\cdot^{\circ}C)\times78.0^{\circ}C=100.0\times4.18\times78.0\ J = 32604\ J\approx32600\ J$.

Answer:

For the first question: D. 2070 J
For the second question: A. 32600 J