Question

question 7 of 15 view policies current attempt in progress a piece of silver whose mass is 10.4 g is immersed in 20.9 g of water. the system is heated electrically from 24.0 to 37.6°c. how many joules of energy are absorbed by the silver? j etextbook and media how many joules of energy are absorbed by the water? j

Explanation:

Part 1: Energy absorbed by silver

Step1: Recall the formula for heat energy

The formula for heat energy absorbed is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature. For silver, the specific heat capacity $c_{\text{Ag}} = 0.235 \, \text{J/g°C}$.

Step2: Calculate $\Delta T$

$\Delta T = T_2 - T_1 = 37.6^\circ\text{C} - 24.0^\circ\text{C} = 13.6^\circ\text{C}$

Step3: Substitute values into the formula

$m_{\text{Ag}} = 10.4 \, \text{g}$, $c_{\text{Ag}} = 0.235 \, \text{J/g°C}$, $\Delta T = 13.6^\circ\text{C}$
$q_{\text{Ag}} = m_{\text{Ag}} \times c_{\text{Ag}} \times \Delta T = 10.4 \, \text{g} \times 0.235 \, \text{J/g°C} \times 13.6^\circ\text{C}$

Step4: Perform the calculation

First, multiply $10.4 \times 0.235 = 2.444$
Then, multiply $2.444 \times 13.6 \approx 33.24$

Step1: Recall the formula for heat energy

Again, use $q = mc\Delta T$. For water, the specific heat capacity $c_{\text{H}_2\text{O}} = 4.184 \, \text{J/g°C}$.

Step2: Use the same $\Delta T$

$\Delta T = 13.6^\circ\text{C}$ (calculated earlier)

Step3: Substitute values into the formula

$m_{\text{H}_2\text{O}} = 20.9 \, \text{g}$, $c_{\text{H}_2\text{O}} = 4.184 \, \text{J/g°C}$, $\Delta T = 13.6^\circ\text{C}$
$q_{\text{H}_2\text{O}} = 20.9 \, \text{g} \times 4.184 \, \text{J/g°C} \times 13.6^\circ\text{C}$

Step4: Perform the calculation

First, multiply $20.9 \times 4.184 \approx 87.4456$
Then, multiply $87.4456 \times 13.6 \approx 1189.26$

Answer:

Approximately $\boldsymbol{33.2}$ J (or more precisely 33.24 J)

Part 2: Energy absorbed by water